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SOLUTIONS
Chapter 1 Solutions
Amido group
⫹H N
3
Ether Carbonyl
OH
HO
C
H2
Hydroxyl
O
CH O
O
CH C
C
C
C
Vitamin C
O
Ether
H3C
H3C
Carbonyl
O
CH3
O
C
H2
Carbonyl
Carboxyl
OH
N
Nicotinic acid (niacin)
OH
HO
O
H
C
C
CH2 H
n
CH3
Coenzyme Q
4. (a) N-acetylglucosamine is a monosaccharide.
(b) CMP is a nucleotide.
(c) Homocysteine is an amino acid.
(d) Cholesteryl ester is a lipid.
6. It is a lipid (it is actually lecithin). It is mostly C and H, with
relatively little O and only one N and one P. It has too little O
to be a carbohydrate, too little N to be a protein, and too little
P to be a nucleic acid.
8. A diet high in protein results in a high urea concentration,
since urea is the body’s method of ridding itself of extra nitrogen.
Nitrogen is found in proteins but is not found in significant
amounts in lipids or carbohydrates. A low-protein diet provides the patient with just enough protein for tissue repair and
growth. In the absence of excess protein consumption, urea
production decreases, and this puts less strain on the patient’s
weakened kidneys.
10. The carbon marked with an asterisk is chiral. This means that
alanine has two possible enantiomers, or mirror-image isomers.
H O
⫹H
3N
C* C
O
O
2. The functional groups are identified below.
O⫺
CH3
12. Two hydrogen atoms and one oxygen atom are lost when
Asn and Cys form a dipeptide. This is an example of a condensation reaction. The carboxylate functional group on the Asn is
lost and the amino functional group on the Cys is lost. An amido
group is formed.
CH C
NH CH C O⫺
CH2
CH2
C
SH
O
NH2
14. (a) Fructose has the same formula, C6H12O6, as glucose.
(b) Fructose is a ketone, whereas glucose is an aldehyde.
16. Nucleotides consist of a five-carbon sugar, a nitrogenous
ring, and one or more phosphate groups linked covalently together.
18. Glucose has several hydroxyl groups and is a polar molecule. As such, it will have difficulty crossing the nonpolar membrane. The 2,4-dinitrophenol molecule consists of a substituted
benzene ring and has greater nonpolar character. Of the two
molecules, the 2,4-dinitrophenol will traverse the membrane
more easily.
20. Polysaccharides serve as fuel-storage molecules and can also
serve as structural support for the cell.
22. The polymeric molecule is more ordered and thus has less
entropy. A mixture of constituent monomers has a large number
of different arrangements (like the balls scattered on a pool table)
and thus has greater entropy.
24. The dissolution of calcium chloride in water is a highly exothermic process, as indicated by the negative value of DH. This means
that when calcium chloride dissolves in water, the system loses heat
to the surroundings and the surroundings become warm. The plastic bag holding the calcium chloride solution becomes warm and
can be used as a hot pack by the hiker at cold temperatures.
26. First, calculate DH and DS, as described in Sample
Calculation 1-1:
DH 5 HB – HA
DH 5 60 kJ ? mol–1 – 54 kJ ? mol–1
DH 5 6 kJ ? mol–1
DS 5 SB – SA
DH 5 43 J ? K–1 ? mol–1 – 22 J ? K–1 ? mol–1
DH 5 21 J ? K–1 ? mol–1
(a) DG 5 (6000 J ? mol–1) – (4 1 273 K)(21 J ? K–1 ? mol–1)
DG 5 180 J ? mol–1
The reaction is not favorable at 48C.
(b) DG 5 (6000 J ? mol–1) – (37 1 273 K)(21 J ? K21 ? mol–1)
DG 5 –510 J ? mol–1
The reaction is favorable at 378C.
28. (a) decrease
(b) increase
CHAPTER 1 Solutions | 1
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30. (a) The reaction is exothermic because the value of DH
is negative.
(b) DG 5 DH 2 TDS
DG 5 28190 J ? mol21 2 (310 K)(2.20 J ? K21 ? mol21)
DG 5 28870 J ? mol21 5 28.87 kJ ? mol–1
The negative value of DG indicates that the reaction is
spontaneous.
(c) The DH term makes a greater contribution to the DG
value. This indicates that the reaction is spontaneous largely
because the reaction is exothermic. The reaction proceeds
without much change in entropy.
32. (a) reduction
(b) oxidation
34. (a) oxidizing agent
(b) oxidizing agent
(c) oxidizing agent
(d) reducing agent
36. The complete oxidation of stearate to CO2 yields more
energy because 17 of the 18 carbons of stearate are fully reduced.
The conversion of these carbons to CO2 provides more free energy than some of the carbons of a-linolenate, which participate
in double bonds and are therefore already partially oxidized.
38. The first biological molecules would have had to polymerize, and they would have had to find some way to make copies
of themselves.
40. It is difficult to envision how a single engulfment event
could have given rise to a stable and heritable association of
the eukaryotic host and the bacterial dependent within a single
generation. It is much more likely that natural selection gradually promoted the interdependence of the cells. Over many generations, genetic information supporting the association would
have become widespread.
Chapter 2 Solutions
2. Because the partial negative charges are arranged symmetrically (and the shape of the molecule is linear), the molecule as a
whole is not polar.
O C O
4. Water has the highest melting point because each water molecule forms hydrogen bonds with four neighboring water molecules, and hydrogen bonds are among the strongest intermolecular forces. Ammonia is also capable of forming hydrogen bonds,
but they are not as strong (due to the electronegativity difference
between hydrogen and nitrogen). Methane cannot form hydrogen bonds; the molecules are attracted to their neighbors only
via weak London dispersion forces.
6. Compound A does not form hydrogen bonds (the molecule
has a hydrogen bond acceptor but no hydrogen bond donor).
Compounds B and C form hydrogen bonds as shown because
each molecule contains at least one hydrogen bond donor and
a hydrogen bond acceptor. The molecules in D do not form
hydrogen bonds with each other, whereas the molecules in E do
2
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because ammonia has a hydrogen bond donor and diethyl ether
has a hydrogen bond acceptor.
(B) N
N
H
N
N
H
(C) H3C
O
H
CH2
O
CH2
H3C
H N H
O
H
H2C
H3C
(E)
H
CH2
CH3
8. (a) H , C , N , O , F
(b) The greater an atom’s electronegativity, the more polar
its bond with H and the greater its ability to act as a hydrogen bond acceptor. Thus, N, O, and F, which have relatively
high electronegativity, can form hydrogen bonds, whereas C,
whose electronegativity is only slightly greater than hydrogen’s, cannot.
10. Aquatic organisms that live in the pond are able to survive
the winter. Since the water at the bottom of the pond remains
in the liquid form instead of freezing, the organisms are able to
move around. The ice on top of the pond also serves as an insulating layer from the cold winter air.
12. The positively charged ammonium ion is surrounded by
a shell of water molecules that are oriented so that their partially negatively charged oxygen atoms interact with the positive
charge on the ammonium ion. Similarly, the negatively charged
sulfate ion is hydrated with water molecules oriented so that the
partially positively charged hydrogen atoms interact with the
negative charge on the sulfate anion. (Not shown in the diagram
is the fact that the ammonium ions outnumber the sulfate ions
by a 2:1 ratio. Also note that the exact number of water molecules shown is unimportant.)
H
H
O
H
O
H
H
O
H
O
H
H
H
O
H
H
O NH
4 O
H
H
H
H
SO42
H
O
H
H
O
O
H
H
14. Structure A depicts a polar compound, while structure B
depicts an ionic compound similar to a salt like sodium chloride.
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This is more consistent with glycine’s physical properties as a
white crystalline solid with a high melting point. While structure A could be water soluble because of its ability to form hydrogen bonds, the high solubility of glycine in water is more
consistent with an ionic compound whose positively and negatively charged groups are hydrated in aqueous solution by water
molecules.
16. (a) A 30X dilution is equivalent to multiplying one-tenth
(1021) by itself 30 times: (1021)30 5 10230. The concentration would be 10230 M.
(b) Use Avogadro’s number and multiply the concentration
by the volume to show that there is much less than one
molecule present in 1 mL:
10.001 L2 110230 moles/L2 16.02 3 1023 molecules/mole2
5 6.02 3 10210 molecules
(c) The ability of water molecules to form a hydrogenbonded coating, or cage, around a solute molecule, particularly a hydrophobic one, might support the idea of water’s
memory. However, water molecules are constantly in motion,
so a group of water molecules that have been in contact with a
solute do not retain an imprint of it.
18. (a) In the nonpolar solvent, AOT’s polar head group faces the
interior of the micelle, and its nonpolar tails face the solvent.
CH2CH3
Nonpolar
tails
O
H3 C
(CH2)3
CH CH2
O
C
CH2 O
H3 C
(CH2)3
CH CH2
O
C
CH S
CH2CH3
O
O
Polar
head
O
AOT
(b) The protein, which contains numerous polar groups, interacts with the polar AOT groups in the micelle interior.
Isooctane
Water
Protein
20. (a) It is doubtful that the contents of the ball could influence the behavior of external water molecules separated by
layers of rubber and plastic.
(b) Even if water clusters were disrupted (which they are not),
the removal of dirt requires more than one individual water
molecule. In order for a dirt molecule to be washed away, it
must be surrounded (solubilized) by many water molecules.
(c) Hot water, because of the higher energy of its water molecules, has intrinsically better dirt-solubilizing power than cold
water, regardless of the presence or absence of detergent. In
the absence of detergent, hot water has significant cleaning
power on its own, which could be attributed to the presence
of a laundry ball.
22. The waxed car is a hydrophobic surface. To minimize its interaction with the hydrophobic molecules (wax), each water drop
minimizes its surface area by becoming a sphere (the geometrical
shape with the lowest possible ratio of surface to volume). Water
does not bead on glass, because the glass presents a hydrophilic
surface with which the water molecules can interact. This allows
the water to spread out.
24. (a) CO2 is nonpolar and would be able to cross a bilayer.
(b) Glucose is polar and would not be able to pass through a
bilayer because the presence of the hydroxyl groups means glucose is highly hydrated and would not be able to pass through
the nonpolar tails of the molecules forming the bilayer.
(c) DNP is nonpolar and would be able to cross a bilayer.
(d) Calcium ions are charged and are, like glucose, highly
hydrated and would not be able to cross a lipid bilayer.
26. The amount of Na1 (atomic weight 23 g ? mol21) lost in
15 minutes, assuming a fluid loss rate of 2 L per hour and a
sweat Na1 concentration of 50 mM, is
0.25 h 3
23 g
1000 mg Na 1
2L
0.05 mol
3
3
3
h
L
mol
g Na 1
1 oz chips
3
5 2.9 oz chips
200 mg Na 1
It would take 2.9 ounces of potato chips (about a handful) to
replace the lost sodium ions.
28. (a) In a high-solute medium, the cytoplasm loses water;
therefore its volume decreases. In a low-solute medium, the
cytoplasm gains water and therefore its volume increases.
(b) E. coli accumulates water when grown in a low-osmolarity
medium. However, regulation of water content only would
cause a large increase in cytoplasmic volume. To avoid this
large increase in volume, E. coli also exports K1 ions. The
opposite occurs when E. coli is grown in a high-osmolarity medium—the cytoplasmic water content is decreased,
but cytoplasmic osmolarity increases as E. coli imports K1
ions. [From Record, M.T., et al., Trends Biochem. Sci. 23,
143–148 (1998).]
30. The HCl is a strong acid and dissociates completely. This
means that the concentration of hydrogen ions contributed by
the HCl is 1.0 3 1029 M. But the concentration of the hydrogen
ions contributed by the dissociation of water is 100-fold greater
than this: 1.0 3 1027 M. The concentration of the hydrogen
ions contributed by the HCl is negligible in comparison. Therefore, the pH of the solution is equal to 7.0.
32.
−
34.
Acid, base, or neutral?
A
B
C
D
acid
base
neutral
acid
pH
[H1] (M)
[OH2] (M)
5.60
7.65
7.00
2.68
2.5 3 1026
2.2 3 1028
1.0 3 1027
2.1 3 1023
4.0 3 1029
4.5 3 1027
1.0 3 1027
4.8 3 10212
CHAPTER 2 Solutions | 3
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36. (a) The final concentration of HCl is
10.0015 L2 13.0 M2
5 0.0045 M
1.0015 L
Since HCl is a strong acid and dissociates completely, the
added [H1] is equal to [HCl]. (The existing hydrogen ion
concentration in the water itself, 1.0 3 1027 M, can be
ignored because it is much smaller than the hydrogen ion
concentration contributed by the hydrochloric acid.)
pH 5 2log [H 1 ]
pH 5 2log10.00452
pH 5 2.3
(b) The final concentration of NaOH is
10.0015 L2 13.0 M2
5 0.0045 M
1.0015 L
Since NaOH dissociates completely, the added [OH2] is
equal to the [NaOH]. (The existing hydroxide ion concentration in the water itself, 1.0 3 1027 M, can be ignored
because it is much smaller than the hydroxide ion concentration contributed by the NaOH.)
Kw 5 1.0 3 10214 5 [H 1 ] [OH 2 ]
1.0 3 10214
[H 1 ] 5
[OH 2 ]
1.0 3 10214
[H 1 ] 5
10.0045 M2
[H 1 ] 5 2.2 3 10212 M
pH 5 2log [H 1 ]
pH 5 2log12.2 3 10212 2
pH 5 11.6
38. The carbonate ions accept protons from water and form
hydroxide ions (as shown in the equation below), resulting in
basic urine.
2
2
CO22
3 1aq2 1 H2O1l 2 S HCO3 1aq2 1 OH 1aq2
40. (a) H2C2O4
(b) H2SO3
(c) H3PO4
(d) H2CO3
(e) H2AsO24
(f ) H2PO24
(g) H2O2
42. (a) CH3
C
O
COO
Pyruvate
(b) The structure of pyruvate will predominate in the cell at
pH 7.4. The pK values for carboxylic acid groups are typically in the 2–3 range; therefore, the carboxylate group will
be unprotonated at physiological pH.
4
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44. The pK of the fluorinated compound would be lower (it is
9.0); that is, the compound becomes less basic and more acidic.
This occurs because the F atom, which is highly electronegative, pulls on the nitrogen’s electrons, loosening its hold on the
proton.
46. (a) 10 mM glycinamide buffer, because its pK is closer to
the desired pH.
(b) 20 mM Tris buffer, because the higher the concentration of the buffering species, the more acid or base it can
neutralize.
(c) Neither; each solution will contain an equilibrium mixture of the boric acid and its conjugate base (borate).
48.
H 1 1aq2 1 HCO32 1aq2 z
y H2CO3 1aq2 z
y H2O1l 2 1 CO2 1aq2
Loss of CO2 through the shell would shift the above equations
to the right. Carbonic acid in the egg would produce more water
and CO2 to make up for the loss of CO2. This in turn would
cause additional hydrogen ions and bicarbonate ions to form
more carbonic acid. The loss of hydrogen ions would result in an
increased pH of the contents of the egg.
50. The aspirin is more likely to be absorbed in the stomach
at pH 2. At this pH, the carboxylate group is mostly protonated and uncharged. This allows the aspirin to pass more easily
through the nonpolar lipid bilayer. At the pH of the small intestine, the carboxylate group is mostly in the ionized form and
will be negatively charged. Charged species are more polar than
uncharged species (and are likely to be hydrated) and will have
difficulty traversing a lipid bilayer.
52. Use the pK value in Table 2-4 and the Henderson–
Hasselbalch equation to calculate the ratio of imidazole (A2) and
the imidazolium ion (HA):
[A2 ]
pH 5 pK 1 log
[HA]
[A 2 ]
log
5 pH 2 pK
[HA]
[A 2 ]
5 101pH2pK 2
[HA]
[A2 ]
5 1017.427.02 5 2.5
[HA]
54. At the final pH,
[ A2 ]
5 101pH2pK 2 5 1015.024.762 5 100.24 5 1.74
[HA]
Adding NaOH to the acetic acid will convert some of the acetic
acid (HA) to acetate (A2):
NaOH 1 CH3COOH z
y Na1 1 CH3COO2 1 H2O
For every mole of NaOH added, one mole of CH3COOH will be
consumed, and one mole of CH3COO2 will be generated.
If x is the number of moles of NaOH added, then x will also be
the number of moles of A2 generated.
Calculate the initial amount of acetic acid:
10.50 L2 3
0.20 mol
5 0.10 mol acetic acid
L
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The initial amount of acetic acid is 0.10 mol, so the final amount
of acetic acid will be 0.10 mol 2 x.
[ A2 ]
x
5 1.74 5
[HA]
0.10 mol 2 x
x 5 1.7410.10 mol 2 x2 5 0.174 mol 2 1.74x
2.74x 5 0.174 mol
x 5 0.174 mol/2.74 5 0.0635 mol
Calculate the mass of NaOH to add:
0.0635 mol
5 1.58 g
40 g ? mol21
56. (a)
CH2OH
HOH2C
NH
3
C
CH2OH
Weak acid
CH2OH
HOH2C
C
NH2
CH2OH
Conjugate base
(b) The pK of Tris is 8.30; therefore, its effective buffering
range is 7.30–9.30.
(c) Rearranging the Henderson–Hasselbalch equation gives
[A 2 ]
5 101pH2pK 2 5 1018.228.32 5 1020.1 5 0.79
[HA]
The final amount of A2 is 44 mmol 1 4.5 mmol 5
48.5 mmol.
The final amount of HA is 56 mmol 2 4.5 mmol 5
51.5 mmol.
Use the Henderson–Hasselbalch equation to calculate the
new pH:
pH 5 pK 1 log
[ A2 ]
48.5
5 8.3 1 log
[HA]
51.5
5 8.3 1 120.0262 5 8.27
The buffer has been effective: The pH has increased only
0.07 unit (from pH 5 8.2 to pH 5 8.27) with the addition of the strong base. In comparison, the addition of
the same amount of base to water, which is not buffered,
resulted in a pH change from approximately 7.0 to 11.6
(see Problem 36b).
58. The ratio of bicarbonate to carbonic acid in the patient’s
blood can be determined using the Henderson–Hasselbalch
equation:
pH 5 pK 1 log
7.55 5 6.35 1 log
Since [A2] 1 [HA] 5 0.1 M, [A2] 5 0.1 M 2 [HA], and
10.1 M 2 [HA] 2
[HA]
0.79 [ HA]
1.79 [ HA]
[HA]
[A2 ] 1 [HA]
5 0.79,
5 0.1 M 2 [ HA]
5 0.10 M
5 10.10 M2/1.79 5 0.056 M 5 56 mM
5 100 mM, so [A2 ] 5 44 mM
(d) When HCl is added, an equivalent amount of Tris
base (A2) is converted to Tris acid (HA). Let x 5 moles of
H1 added 5 (0.0015 L)(3.0 mol ? L21) 5 0.0045 moles 5
4.5 mmol.
The final amount of A2 is 44 mmol 2 4.5 mmol 5 39.5
mmol.
The final amount of HA is 56 mmol 1 4.5 mmol 5 60.5
mmol.
Use the Henderson–Hasselbalch equation to calculate the
new pH:
pH 5 pK 1 log
[A 2 ]
39.5
5 8.3 1 log
[HA]
60.5
5 8.3 1 120.182 5 8.12
The buffer has been effective: The pH has declined about
0.1 unit (from pH 5 8.2 to pH 5 8.1) with the addition of the strong acid. In comparison, the addition of
the same amount of acid to water, which is not buffered,
resulted in a pH change from approximately 7.0 to 2.35
(see Problem 36a).
(e) When NaOH is added, an equivalent amount of Tris
acid (HA) is converted to Tris base (A2). Let x 5 moles of
OH2 added 5 (0.0015 L)(3.0 mol ? L21) 5 0.0045 moles 5
4.5 mmol.
[A2 ]
[HA]
101.2 5
[HCO32 ]
[H2CO3 ]
[HCO32 ]
[H2CO3 ]
[HCO32 ]
15.8
5
[H2CO3 ]
1
pH 5 8.3 1 121.32 5 7.0
Similarly, the ratio of bicarbonate to carbonic acid in a normal
person’s blood can be determined:
[A2 ]
[HA]
[HCO32 ]
7.4 5 6.35 1 log
[H2CO3 ]
2
[HCO3 ]
101.05 5
[H2CO3 ]
2
[HCO3 ]
11.2
5
[H2CO3 ]
1
pH 5 pK 1 log
In order to serve as an effective buffer (i.e., absorb both added
H1 and OH2), both a conjugate base and a weak acid must be
present. In the patient, the ratio of conjugate base to weak acid
does not lie within an effective buffering range. The bicarbonate
concentration (conjugate base) is too high relative to the carbonic acid (weak acid) concentration; thus the relative amount
of weak acid is insufficient.
Chapter 3 Solutions
2. These experiments showed that the transforming factor was
neither a protein nor RNA.
CHAPTER 3 Solutions | 5
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4. The triple-helical model is not consistent with the hydrophobic effect, which suggests that the nonpolar nitrogenous
bases would reside in the center of the DNA structure and
the hydrophilic phosphates would reside on the surface. The
triple-helical model also assumes that the phosphate groups are
protonated and form stabilizing hydrogen bonds in the DNA
interior. But the pK value for phosphate is well below 7, so
the phosphate groups would not be protonated at physiological
pH. In the absence of hydrogen bonds, there are no additional
forces that would hold the strands of the triple helix together.
31% G, since [C] 5 [G]). Each cell is a diploid, containing
60,000 kb or 6 3 107 bases. Therefore,
[A] 5 [T] 5 (0.19)(6 3 107 bases) 5 1.14 3 107 bases
[C] 5 [G] 5 (0.31)(6 3 107 bases) 5 1.86 3 107 bases
12.
N
H
N
6. (a)
NH
O
H
H
N
N
N
Hypoxanthine
O
C
O
N
H
NH2 Base
N
Cytosine
CH2 O
O
H
H
H
H
HO
N
Monosaccharide
H
OH
N
O
P
P
N
N
O
O
N
O
CH2
N
Base
H
H
H
HO
Monosaccharide
O
O
C
OH
H
C
OH
C
H3C
O
O
P
O
H
Base
N
N
H
H
OH
Monosaccharide
OH
O
N
Base
O
(b) A diphosphate bridge links the ribose groups in each
dinucleotide. This linkage is a variation of the monophosphate bridge (phosphodiester linkage) in DNA and
RNA.
(c) The adenosine group in CoA bears a phosphoryl group
on C39.
O
O
P
O
H
H
O
O
H
NH
N
N
Hypoxanthine
16. The DNA contains 50% G 1 C, so its melting point would
be approximately 908C.
110
100
H
H
OH
10. The organism must also contain 19% A (since [A] 5 [T]
according to Chargaff ’s rules) and 62% C 1 G (or 31% C and
6
O
14. It is certainly the case that hydrogen bonds hold A:T and
G:C base pairs together and that these interactions are very favorable. But upon denaturation of the DNA, each nitrogenous
base has the opportunity to form equally favorable hydrogen
bonds with water. Therefore, forces other than hydrogen bonds
must contribute to the overall stability of the DNA molecule.
O
H2
C
H
Uracil
| CHAPTER 3 Solutions
Tm (°C)
8.
N
H
Monosaccharide
NH
N
O
N
N
OCH2 O
H
N
N
H3C
P
OH
CH2
Hypoxanthine
OH
N
H
N
N
N
N
H
Adenine
NH2
CH2 O
H
N
H
H
N
NH2
O
O
O
NH
O
H
90
80
70
0
10
20
30
40
50
GC content (%)
60
70
80
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18. The positively charged sodium ions can form ion pairs with
the negatively charged phosphate groups on the DNA backbone
and “shield” the negative charges from one another. This increases
the overall stability of DNA and makes it more difficult to melt.
20. The positively charged side chains of the Lys and Arg
residues form ion pairs with the negatively charged phosphate
groups on the DNA backbone. These are strong interactions, so
the histones have a high affinity for DNA.
22. The DNA isolated after one generation is a homogeneous
sample of DNA with a density intermediate between DNA containing all 14N and all 15N. The DNA isolated after the second
generation is heterogeneous. Half of the DNA has the same density as the first generation; half of the DNA consists of all 14N
DNA and has a lower density.
15N
14N
First-generation
daughter molecules
Second-generation
daughter molecules
24. The number of possible sequences of four different nucleotides
is 4n where n is the number of nucleotides in the sequence. Therefore: (a) 41 5 4, (b) 42 5 16, (c) 43 5 64, and (d) 44 5 256.
26. The genetic code (shown in Table 3-3) is redundant. Since there
are 64 different possibilities for 3-base codons and only 20 amino
acids, some amino acids have more than one codon. If a mutation
just happens to occur in the third position (39 end), the mutation
might not alter the protein sequence. For example, GUU, GUC,
GUA, and GUG all code for valine. A mutation in the third position of a valine codon would still result in the selection of valine and
would have no effect on the amino acid sequence of the protein.
28. (a) The siRNA is an “antisense” mRNA, so its sequence
should be complementary to that of the mRNA. The solution below shows an siRNA that corresponds to the 59 end
of the mRNA:
mRNA 59 S 39 AUG GGC UCC AUC GGA GCA GCA AGC AUG GAA
siRNA 39 S 59 UAC CCG AGG UAG CCA CGU CGU UCG UAC CUU
(b) When the antisense RNA binds to the mRNA, the
mRNA can no longer serve as a template for protein
synthesis. (In the cell, the binding of antisense RNA to
mRNA involves an enzyme that degrades the mRNA
into smaller fragments.)
(c) The siRNA molecules must be able to cross the cell
membrane to enter the cell and find the target mRNA.
30. Prokaryotes tend to have smaller genomes than eukaryotes,
so evolution has shaped the prokaryote genome to pack in genes
more efficiently. Eukaryotes, with larger genomes and more noncoding DNA, have more space to arrange genes.
32. Questions that need to be addressed in order to solve the
C-value paradox:
• How does one define organismal complexity? Perhaps humans are not the most complex organisms.
• How much of the organism’s genome codes for RNA? For
protein? What are the biological roles of these gene products?
• How many copies of each gene are in the organism’s genome?
• What is the amount of noncoding DNA in the genome?
Is the noncoding DNA really “junk DNA,” or does it
have some biological role?
• How many transposable elements are in the noncoding
DNA?
34. The polymerization reaction must be carried out at high
temperatures (hence the need for a heat-stable DNA polymerase)
in order to ensure that the template DNA remains an unknotted
single strand. The high temperature is necessary to melt GC-rich
DNA, which is more stable than AT-rich DNA.
36. To amplify the protein-coding DNA sequence, the primers should correspond to the first three and last three residues of
the protein (each amino acid represents three nucleotides, so the
primers would each be nine bases long). Use Table 3-3 to find the
codons that correspond to the first three residues:
Met
AUG
Gly
GGU
GGC
GGA
GGG
Ser
UCU
UCC
UCA
UCG
AGU
AGC
Using just the topmost set of codons, a possible DNA primer
would therefore have the sequence 59-ATGGGTTCT-39. This
primer could base pair with the gene’s noncoding strand, and its
extension from its 39 end would yield a copy of the coding strand
of the gene (see Fig. 3-18). The other primer must correspond to
the last three amino acids of the protein:
Val
GUU
GUC
GUA
GUG
Ser
UCU
UCC
UCA
UCG
AGU
AGC
Pro
CCU
CCC
CCA
CCG
Again, considering just the topmost set of codons, a probable DNA
coding sequence would be 59-GTTTCTCCT-39. This sequence
cannot be used as a primer. However, a suitable primer would be
the complementary sequence 59-AGGAGAAAC-39, which can
then be extended from its 39 end to yield a copy of the noncoding
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strand of the gene. The number of possible primer pairs is quite
large, because all but one of the amino acids have more than one
codon. For the first primer, there are 1 3 4 3 6 5 24 possibilities;
for the second, 4 3 6 3 4 5 96 possibilities. There are 24 3 96
5 2304 different pairs of primers that could be used to amplify
the gene by PCR.
38. The enzyme MspI generates sticky ends. The single-stranded
regions are then removed by the action of the exonuclease, releasing the free nucleotides C and G.
Exonuclease
cleavage
6. The polypeptide would be even less soluble than free Tyr,
because the amino and carboxylate groups that interact with
water and make Tyr soluble are lost in forming the peptide bonds
in poly(Tyr).
8. Histones contain an abundance of the positively charged
amino acids lysine and arginine. The positive charges of these
amino acid side chains interact electrostatically to form ion pairs
with the negatively charged phosphate groups on the backbone
of the DNA molecule and to minimize charge–charge repulsion
of the negatively charged phosphate groups.
10. Threonine has two chiral carbons; therefore, four stereoisomers are possible.
COO
TG C T TA G C
A C G A ATC G G C
40.
CGGAACGA
C T TG C T
EcoRI
Pst I
2 kb
7 kb
2. If amino acids existed in the nonionic form shown in Problem 1, they would have lower melting points and would be
soluble in nonpolar organic solvents. High melting points and
water solubility are characteristics of ionic substances. These
observations support the zwitterionic (doubly ionic) form of
the amino acid.
4.
O
O
CH C
O
H3
O
NH
4
H
3N
CH2
C
O
O
CH2
H
CH C
(CH2)2
C
H
O
NH2
Glutamine
8
| CHAPTER 4 Solutions
OH
NH
4
C
NH
3
HO
C
H
CH3
CH3
COO
COO
3N
C
H
H
C
NH
3
HO
C
H
H
C
OH
O
O
H N
3
-Amino group
CH C
CH2
CH2
CH2
CH2
CH2
C
CH2
O
NH
3
-Amino group
CH
HN
C
O
C-Terminus
-Carboxylate group
O
-Carboxylate group
Lys-Glu
14.
H
3N
CH
C
O
N
H
CH2
CH C
HN CH
O
O
C
HN CH C
CH2
H
H
O
N
H
CH C
O
CH2
CH2
S
Met-enkephalin
CH3
OH
O
CH C
(CH2)2
H
H
3N
CH
CH2
C
O
N
H
CH C
H
HN CH
H
O
O
C
HN CH C
CH2
O
N
H
CH C
CH2
HC
O
O
Glutamate
CH3
Peptide bond
O
O
C
H3N
12.
O
O
H3O
C
H
CH3
O
Aspartate
O
3N
H
O
CH C
C
NH2
Asparagine
H
H
N-Terminus
Chapter 4 Solutions
3N
C
3 kb
42. To design an oligonucleotide probe for a gene, the researcher
must apply the genetic code in reverse, that is, select codons
that correspond to the amino acids in the protein. Most amino
acids can be encoded by more than one codon, so the researcher
would have to choose one of them and hope that it matched
the DNA well enough for the probe to successfully hybridize
with the DNA. Met and Trp, however, are encoded by only one
codon each, so by using these codons as part of the probe, the
researcher can be assured of a perfect match with the DNA, at
least for these three nucleotides.
H
H3N
Pst I
8 kb
COO
Leu-enkephalin
OH
CH3
CH3
O
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16. A polypeptide is a single polymer of amino acids. A protein
may consist of one or more polypeptide chains.
18. The amino group of Pro is linked to its side chain (see Fig.
4-2), which limits the conformational flexibility of a peptide
bond involving the amino group. The geometry of this peptide
bond is incompatible with the bond angles required for a polypeptide to form an a helix.
20.
G
F
L
S
L
A
L
K
A
S
W
I
K
A
P
G
L
F
The polar amino acid residues are shown in red; the nonpolar residues are shown in blue. This is another example of an
amphipathic helix (see Problem 19). The hydrophobic side of
the helix will interact favorably with the nonpolar membrane
and disrupt its structure, resulting in cell lysis. (The investigators of the study found that the peptide was able to form a pore
in the membrane, disrupting the ionic balance and killing the
cell.) [From Corzo, G., Escoubas, P., Villegas, E., Barnham, K. J.,
Weilan, H. E., Norton, R. S., and Nakajima, T., Biochem. J. 359,
35–45 (2001).]
22. (a) His (b) Ser (c) Tyr (d) Cys (e) Asn
24. (a) Phe. Ala and Phe are both hydrophobic, but Phe is much
larger and might not fit as well in Val’s place.
(b) Asp. Replacing a positively charged Lys residue with an
oppositely charged Asp residue would be more disruptive.
(c) Glu. The amide-containing Asn would be a better substitute for Gln than the acidic Glu.
(d) His. The geometry of a Pro residue constrains the conformation of the polypeptide. Gly, which lacks a side chain,
can adopt the same backbone conformation, but a residue
with a bulkier side chain cannot.
26. Substitution of a histidine for an arginine evidently causes
a change in the three-dimensional structure of the protein,
which adversely affects its function. This could occur for
a variety of reasons. Histidine’s side chain is composed of a
five-membered ring, whereas arginine’s side chain is a straightchain structure. The change in shape of the side chain could
lead to an overall change in the shape of the protein. Because
of the difference in their structures (although both amino acids
can form hydrogen bonds and ion pairs), the substituted histidine might not form an interaction that is crucial to the proper
functioning of the protein. A “permanent” positive charge
might also be necessary. Arginine has a pK of 12.5 and thus
is always protonated at physiological pH. Even though the pK
value of an amino acid in a protein is not necessarily the same
as the pK value of the free amino acid, the great difference in
the pK values indicates that the arginine is far more likely to be
protonated than the histidine. If a full strong charge at this site
is necessary for protein function, its replacement could result
in a defective protein.
28. (a) Heating causes a protein to “melt,” or unfold, because
heating increases the vibrational and rotational energy of
atoms in the protein, which disrupts the weak interactions
that keep the protein in its properly folded state.
(b) pH changes alter the ionization states of amino acid side
chains. This affects the ability of side chains to form ion
pairs. Hydrogen bonds may also be broken as protonation
or deprotonation renders amino acid side chains unable to
serve as hydrogen bond donors or acceptors.
(c) Detergents have a nonpolar domain that allows them
to penetrate into the interior of the protein, thus interfering with the hydrophobic interactions responsible for the
protein’s tertiary structure.
(d) Reducing agents, such as 2-mercaptoethanol, break
disulfide bonds (converting them to the —SH form) and
destabilize proteins that require disulfide bonds in order to
assume the correct conformation.
30. The loss of the C chain means the loss of information that
is essential to the proper folding of insulin. Removal of the C
chain leaves two separate chains (A and B), and it is much more
difficult for two chains to resume their native conformation than
for one chain to do so. Proinsulin has no difficulty resuming its
native conformation in a denaturation/renaturation experiment
because proinsulin consists of only one peptide chain. When
proinsulin is converted to insulin, two peptide bonds are cleaved
to produce the correctly cross-linked A and B chains.
32. Proteins with a higher OSH molar content have a greater
opportunity to form disulfide bonds, which play a role in stabilizing the protein. Proteins strengthened with disulfide bonds
would require higher temperatures to denature and would
therefore have higher Tm values. [From Lacy, E. R., Baker, M.,
and Brigham-Burke, M., Anal. Biochem. 382, 66–68 (2008).]
34. Protein loops are often at the protein surface, whereas regular secondary structure predominates in the protein core. The
loops are better able to accommodate changes in size and amino
acid composition than the core segments, where a change in size
or composition may disrupt an a helix or b sheet that is an essential part of the protein’s structure.
36. Proteins and their component amino acids (and many other
organic compounds) have inherent “handedness” and therefore
cannot be interconverted through mirroring. Consider a righthanded a helix; its mirror image would be a left-handed a helix,
which is not found in nature.
38. Because the tetramers dissociate into dimers in the presence of the amphipathic SDS, it is possible that hydrophobic interactions are important in stabilizing the dimer. The
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hydrophobic surface of the dimer could interact with the nonpolar tail of SDS when the tetramer dissociates into dimers.
But the monomers might be stabilized by a different type of
interaction, either polar or ionic. The nonpolar tail of SDS
would not interact favorably with a polar surface, and the
negatively charged head group of SDS would be repelled by
negatively charged amino acid side chains involved in ionic interactions. The association of the monomers with one another
is more favorable than their association with SDS would be, so
the dimers resist dissociation into monomers in the presence of
detergent. [From Gentile, F., Amodeo, P., Febbraio, F., Picaro,
F., Motta, A., Formisano, S., and Nucci, R., J. Biol. Chem.,
277, 44050–44060 (2002).]
Chymotrypsin fragments
HSEGTF
SNDY
SKY
LEDRKAQDF
VRW
LMNNKRSGAAE
40. (a)
HSEGTFSNDYSKYLEDRKAQDFVRWLMNNKRSGAAE
O
H
N
CH
O
H
N
C
CH2
DTT
O
H
N
C
CH2
S
HN
CH
CH
C
CH2
ICH2COOH
SH
S
CH2COO⫺
CH2COO⫺
S
SH
S
CH2
CH2
CH2
CH
C
HN
CH
O
Trypsin fragments
HSEGTFSNDYSK
YLEDRK
AQDFVRWLMNNK
RSGAAE
C
HN
O
CH
44. Thermolysin would yield the most fragments (nine) and
chymotrypsin would yield the fewest (three).
46. (a) Since each codon corresponds to an amino acid, the
error rate is
5 3 1024 error
3 500 residues 5 0.25
residue
C
O
About one-quarter of the polypeptides would contain a
substitution.
(b) Virtually all the 2000-residue polypeptides would contain a substitution:
Absorbance @ 280 nm
(b) A simplified version of the anion exchange chromatogram is shown.
Ara h8
Ara h6
high salt
5 3 1024 error
3 2000 residues 5 1.0
residue
48. Because the Edman degradation reaction requires a free
amino group (see Fig. 4-22), mass spectrometry should be used
to determine the polypeptide’s sequence.
50.
H
(c) Treatment of the proteins with a reducing agent followed by iodoacetic acid converts five disulfide bridges
(which are neutral) in Ara h6 to ten negatively charged side
chains. The treatment produces a modified Ara h6 protein
with a more acidic pI. The two proteins were successfully
separated because the Ara h8 protein is less attracted to the
positively charged anion exchange beads and elutes sooner.
The Ara h6 binds more strongly to the anion exchange
beads and elutes only after salt has been added to the elution buffer. [From Riecken, S., Lindner, B., Petersen, A.,
Jappe, U., and Becker, W.-M., Biol. Chem., 389, 415–423
(2008).]
42. The cleavage site for each fragment is highlighted. A fragment not ending in Phe, Tyr, or Trp (for chymotrypsin cleavage)
or Lys or Arg (for trypsin cleavage) must be the carboxyl terminal
fragment. Use “overlap” to work backward from the C-terminus
to determine the sequence of the polypeptide.
10
| CHAPTER 5 Solutions
N
O
Solvent volume
H
H
C
H
H
C
H
H
H
C
H
C
C
H
H C
C
C
N
C
N
N
C
C
H
H
O
H
C
O
H
H
H
S
C
H
H
H
O
H
Chapter 5 Solutions
2.
H
Y5
Y5
1 pO2 2 n
1 p50 2 n 1 1 pO2 2 n
125 torr2 3
115 torr2 3 1 125 torr2 3
Y 5 0.82
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Y5
Y5
1 pO2 2 n
1 p50 2 1 1 pO2 2
n
n
1120 torr2 3
115 torr2 3 1 1120 torr2 3
Y 5 1.00
4. As shown in Solution 2, hemoglobin is completely saturated
with oxygen (Y 5 1.00) when the p O2 5 120 torr. When the
p O2 decreases to 25 torr, Y 5 0.82. This means that 18% of
oxygen bound to hemoglobin is delivered to the tissues when
p CO2 5 5 torr. But when the p CO2 increases to 40 torr, the
amount of oxygen delivered to tissues increases dramatically, as
shown in Solution 3. Under these conditions, hemoglobin is
nearly saturated with oxygen when p O2 5 120 torr (Y 5 0.96).
But when p O2 is 25 torr in the tissues, Y decreases to 0.20. So
76% of the oxygen is delivered to tissues. In general, higher CO2
concentrations assist hemoglobin in delivering oxygen from the
lungs to the tissues. [From Bohr, C., Hasselbalch, K., and Krogh,
A., Skand. Arch. Physiol. 16, 401–412 (1904).]
6. Globin lacks an oxygen-binding group and therefore cannot
bind O2. Heme alone is oxidized and therefore cannot bind O2.
The bound heme gives a protein such as myoglobin the ability
to bind O2. In turn, the protein helps prevent oxidation of the
heme Fe atom.
8. His F8 is the most likely to be invariant. The nitrogen in the
imidazole ring of His F8 serves as one of the ligands to the iron
in the heme group. This residue plays a critical role in the structures of myoglobin and hemoglobin and is essential for the proper
binding of oxygen; substitution of this amino acid would likely
interfere with the ability of the protein to bind and release oxygen
effectively.
10. (a) Vitamin O is useless because the body’s capacity to
absorb oxygen is not limited by the amount of oxygen
available but by the ability of hemoglobin to bind and
transport O2. Furthermore, oxygen is normally introduced
into the body via the lungs, so it is unlikely that the gastrointestinal tract would have an efficient mechanism for
extracting oxygen.
(b) The fact that oxygen delivery in vertebrates requires a
dedicated O2-binding protein (hemoglobin) indicates that
dissolved oxygen by itself cannot attain the high concentrations required. (In fact, the solubility of oxygen in pure water is only about 0.1 mM; the hemoglobin in blood boosts
the solubility to about 10 mM.) In addition, a few drops of
vitamin O would make an insignificant contribution to the
amount of oxygen already present in a much larger volume
of blood.
12. A decrease in pH diminishes hemoglobin’s affinity for oxygen (the Bohr effect), thereby favoring deoxyhemoglobin. Since
only deoxyhemoglobin S polymerizes, sickling of cells is most
likely to occur when the parasite-induced drop in pH promotes
the formation of deoxyhemoglobin.
14. The p50 for hemoglobin when p CO2 5 5 torr is given as
15 torr, whereas the p 50 when p CO2 5 80 torr is given as 40 torr.
When the partial pressure of CO2 increases, the p 50 for hemoglobin
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increases as well. An increased p 50 indicates a lower affinity of hemoglobin for oxygen. Therefore, when the concentration of CO2
increases, as it does in metabolically active tissues, hemoglobin’s affinity for oxygen decreases and hemoglobin unloads its oxygen to
the cells that need it.
16. (a) HCO 32 is formed when carbon dioxide reacts with water
to form carbonic acid, a reaction in the blood catalyzed by
carbonic anhydrase. The carbonic acid dissociates to form
protons and HCO 32 .
(b) Both curves are sigmoidal. According to the investigators, the p50 for crocodile hemoglobin is 6.8 torr in the
absence of bicarbonate and 44 torr in the presence of bicarbonate. The higher p 50 value in the presence of bicarbonate
means that the hemoglobin has a lower affinity for oxygen.
w/o HCO3
Y
w/ HCO3
p O2
(c) The lysine side chain is positively charged and can interact with the negative charge of the bicarbonate. The phenolate oxygens on the two tyrosine side chains can act as
hydrogen bond acceptors for the bicarbonate hydrogen.
(d) All of the allosteric effectors are negatively charged.
They bind to hemoglobin by forming ion pairs with positively charged amino acid side chains (such as Lys or Arg)
or with the positively charged a-amino groups at the amino
termini of the four polypeptide chains. [From Komiyama,
N.H., Miyazaki, G., Tame, J, and Nagai, K., Nature 373,
244–246 (1995).]
18. Hydroxyurea increases the patient’s blood concentration of
hemoglobin F, which has two a chains and two g chains rather
than two a chains and two b chains. Since the b chain has been
mutated in the patient suffering from sickle cell anemia, the
increased synthesis of hemoglobin F allows the patient to use
hemoglobin F to transport oxygen instead of the defective hemoglobin S. In effect, the defective b chains are replaced with
healthy g chains.
20. Hb Bruxelles is the most unstable, since Phe is deleted. Hb
Hammersmith is next, since the nonpolar Phe has been mutated
to a polar Ser. Hb Sendagi and Hb Bucuresti have nonpolar
amino acid substitutions, but Hb Bucuresti is more stable than
Hb Sendagi because Leu is larger than Val and thus closer in size
to the large nonpolar Phe. [From Griffon, N., Baden, C., LenaRusso, D., Kister, J., Barkakdjian, J., Wajcman, H., Marden, M.C.,
and Poyart, C., J. Biol. Chem. 271, 25916–25920 (1996).]
22. (a) Keratin and collagen; (b) myosin and kinesin; (c) actin
and tubulin; (d) actin, myosin, tubulin, and kinesin.
24. (a) Adding another protein was necessary because G-actin
in solution tends to polymerize rather than form crystals.
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The added protein bound to actin and prevented its polymerization.
(b) The crystal structure of the protein alone was needed
so that it could be “subtracted” from the crystal structure of
the actin–protein complex.
26. Because each protofilament contains many tubulin dimers,
the separation of protofilaments during fraying represents a
greater loss of tubulin subunits than if the tubulin subunits left
the microtubule one dimer at a time.
28. Although the drugs have opposite effects on microtubule
dynamics, they both interfere with the normal formation of the
mitotic spindle, which is required for cell division.
30. Paclitaxel increases the stability of the microtubule if it is
able to override the effects of GTP hydrolysis, since it is GTP
hydrolysis to GDP that results in a curved protofilament that
causes the microtubule to have a “frayed” appearance and increases the likelihood that the tubulin dimers in the protofilament will dissociate. [From Amos, L.A., and Löwe, J., Chem.
Biol. 6, R65–R69 (1999).]
32. Microtubules in the cell spindle are dynamic structures since
the mitotic spindle must form during mitosis and degrade after cell
division has occurred. Therefore, these microtubule structures are
less stable than those that make up the structures of the axons of
nerve cells. Axonal microtubules are more stable because of their
structural role.
34. A fibrous protein such as keratin does not have a discrete
globular core. Most of the residues in its coiled-coil structure
are exposed to the solvent. The exception is the strip of nonpolar side chains at the interface of the two coils.
36. The reducing agent breaks the disulfide bonds (—S—S—)
between keratin molecules. Setting the hair brings the reduced
Cys residues (with their —SH groups) closer to new partners on
other keratin chains. When the hair is then exposed to an oxidizing agent, new disulfide bonds form between the Cys residues and
the hair retains the shape of the rollers.
38. The enzyme degrades collagen (which has a repetitive
Gly–X–Y sequence, with X often being Pro). Since collagen is
the major protein present in connective tissue, degradation of
this tissue would facilitate the invasion of the bacterium into the
host. The bacterium itself is not affected because bacteria do not
contain collagen.
40. (a) Collagen B is from rat, and collagen A is from the sea
urchin.
(b) The stability of each of these collagens is correlated
with their imino acid content. The higher the percentage of hydroxyproline, the more regular the structure and
the more difficult it is to melt, resulting in more stable
collagen. The rat has a more stable collagen, and the sea
urchin, which lives in cold water, has a less stable collagen. It is important to note that the melting temperatures
of each collagen molecule are higher than the temperature at which each organism lives. Thus, each organism
has stable collagen at the temperature of its environment.
[From Mayne, J., and Robinson, J. J., J. Cell. Biochem. 84,
567–574 (2001).]
12
| CHAPTER 5 Solutions
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42. (a) (Pro–Pro–Gly)10 has a melting temperature of 418C,
while (Pro–Hyp–Gly)10 has a melting temperature of 608C.
(Pro–Hyp–Gly)10 and (Pro–Pro–Gly)10 both have an imino acid content of 67%, but (Pro–Hyp–Gly)10 contains
hydroxyproline, whereas (Pro–Pro–Gly)10 does not. Hydroxyproline therefore has a stabilizing effect relative to proline.
(b) (Pro–Pro–Gly)10 and (Gly–Pro–Thr(Gal))10 have the
same melting point, indicating that they have equal stabilities. This is interesting because (Pro–Pro–Gly)10 has an
imino acid content of 67%, whereas (Gly–Pro–Thr(Gal))10
has an imino acid content of only 33%. The glycosylated
threonine must have an effect similar to that of proline. It is
possible that the galactose, which contains many hydroxyl
groups, provides additional sites for hydrogen bonding and
would thus contribute to the stability of the triple helix.
(c) The inclusion of (Gly–Pro–Thr)10 is important because
the results show that this molecule doesn’t form a triple helix. This molecule is included as a control to show that the
increased stability of the (Gly–Pro–Thr(Gal))10 is due to the
galactose, not to the threonine residue itself. [From Bann,
J. G., Peyton, D. H., and Bächinger, H. P., FEBS Lett. 473,
237–240 (2000).]
44. Because collagen has such an unusual amino acid composition (almost two-thirds consists of Gly and Pro or Pro derivatives), it contains relatively fewer of the other amino acids and
is therefore not as good a source of amino acids as proteins containing a greater variety of amino acids.
46. The individual collagen chains are synthesized on the ribosome from the constituent amino acids. Proline is one of the
20 amino acids for which a codon exists on the DNA. Some of
the prolines are modified posttranslationally (after the protein is
synthesized) to hydroxyproline. Individual hydroxyproline amino
acids are not incorporated into the protein during synthesis
(there is no codon for hydroxyproline). So [14C]-hydroxyproline
is not used in the synthesis of collagen and no radioactivity appears in the collagen product.
48. (a) The open reading frame is shown below. Collagen has the
structure (Gly–X–Y)n , where X is often proline and Y is often
hydroxyproline. The codon for Gly is GGX, and the codon
for Pro is CCX (where X represents any nucleotide). In the
mutant polypeptide, a Gly residue has been replaced by Asp.
Normal a2(I) collagen gene
. . . CT|GGT|GCT|GTT|GGC|CCA|AGA|GGT|CCT|AGT|GGC|CCA|C. . .
. . . Gly Ala Val Gly Pro Arg Gly Pro Ser Gly Pro . . .
Mutant a2(I) collagen gene
. . . CT|GGT|GCT|GTT|GGC|CCA|AGA|GAT|CCT|AGT|GGC|CCA|C . . .
. . . Gly Ala Val Gly Pro Arg Asp Pro Ser Gly Pro . . .
(b) The Tm value for the mutant collagen would be lower than
the Tm for normal collagen. The substitution of an Asp for a
Gly would disrupt the triple helix in this region of the molecule.
In order for the three chains to pack together to form the triple
helix, every third residue must be a Gly. There are three amino
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acids per turn; therefore, the side chain of the Gly ends up on
the interior of the triple helix. There is not sufficient room to
accommodate a larger side chain of an amino acid such as Asp.
The actual Tm values for normal collagen and the mutant collagen presented here are 418C and 398C, respectively.
(c) The patient suffered from osteogenesis imperfecta, a disease characterized by amino acid changes in type I collagen,
which is found mainly in bones and tendons. The patient’s
bones and tendons could not form properly due to the defects
in the structure of collagen, and the patient died as a result.
[From Baldwin, C.T., Constantinou, C.D., Dumars, K.W.,
and Prockop, D.J., J. Biol. Chem. 264, 3002–3006 (1989).]
50. For many proteins, bacterial expression systems offer a convenient source of protein for structural studies. However, even if
collagen genes were successfully introduced into bacterial cells,
the cells would not be able to produce mature collagen molecules
because collagen is processed after it is synthesized. Bacteria are
unable to undertake some of the processing steps, such as cleavage by extracellular proteases and covalent modification of Pro
and Lys residues.
52.
Cargo
ADP
actin
ATP
ATP binding to the trailing head induces
a conformational change that causes it
to release actin.
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[From Walker, M.L., Burgess, S.A., Sellers, J.R., Wang, F.,
Hammer, J.A., III, Trinick, J., and Knight, P. J., Nature 405,
804–807 (2000).]
54. Myosin’s two heads act independently, so only one binds
to the actin filament at a time. Consequently, when myosin
advances from one actin subunit to another, the protein dissociates completely from its track, much like a one-legged hop.
In contrast, kinesin’s two heads work together, so one head
remains bound to the microtubule track when the other is
released. This mechanism is similar to a two-legged walk.
Chapter 6 Solutions
2. Myosin and kinesin are enzymes because they catalyze the
hydrolysis of ATP. For myosin, the reaction is
ATP 1 H2O S ADP 1 Pi
For kinesin, the reaction is
2 ATP 1 2 H2O S 2 ADP 1 2 Pi
Two cycles of ATP hydrolysis are required to restore the twoheaded kinesin motor to its original position.
4. As shown in Table 6-1, the only relationship between the rate
of catalyzed and uncatalyzed reactions is that the catalyzed reaction is faster than the uncatalyzed reaction. The absolute rate
of an uncatalyzed reaction does not correlate with the degree to
which it is accelerated by an enzyme.
6. (a) isomerase
(b) lyase
(c) oxidoreductase
(d) hydrolase
8. Malate dehydrogenase is an oxidoreductase.
COO
CH2
ATP
ADP
CH
Hydrolysis of ATP to ADP + Pi triggers
a conformational change that swings the
trailing head forward. This also increases
the affinity of the head for actin.
ADP Pi
ADP
ADP
The new leading head binds to the actin
filament. This causes the release of the
ADP from the new trailing head.
ADP Pi
Pi
Pi dissociates from the leading head,
preparing the myosin for another
reaction cycle.
OH
COO
Malate
CH2
malate dehydrogenase
C
COO
O
COO
Oxaloacetate
10. (a) glucose-6-phosphate dehydrogenase
(b) isocitrate lyase
(c) phosphoglycerate kinase
(d) pyruvate carboxylase
12. The common name is argininosuccinate lyase.
14. At first, the temperature increases the reaction rate because
heat increases the proportion of reacting groups that can achieve
the transition state in a given time. However, when the temperature rises above a certain point, the heat causes the enzyme,
which is a protein, to denature. Because most proteins are only
marginally stable (see Section 4-3), denaturation occurs readily,
accounting for the steep drop in enzymatic activity.
16. Yes. An enzyme decreases the activation energy barrier for
both the forward and the reverse directions of a reaction.
18. (a) Gly, Ala, and Val have side chains that lack the functional groups required for acid–base or covalent catalysis.
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(b) Since benzoate resembles the substrate, it is likely that
benzoate binds to the active site of the enzyme. Under
these conditions, FDNP does not have access to the active
site and will be unable to react with the tyrosine (which
is also assumed to be part of the enzyme’s active site because of its unusual reactivity). [From Nishino, T., Massey,
V., and Williams, C.H., J. Biol. Chem., 255, 3610–3616
(1979).]
(b) Mutating one of these residues may alter the conformation at the active site enough to disrupt the arrangement of
other groups that are involved in catalysis.
20. (a) In order for any molecule to act as an enzyme, it must
be able to recognize and bind a substrate specifically, it must
have the appropriate functional groups to effect a chemical
reaction, and it must be able to position those groups for
reaction.
(b) DNA, as a double-stranded molecule, has limited conformational freedom. RNA, which is single-stranded, is able
to assume a greater range of conformations. This flexibility
allows it to bind to substrates and carry out chemical transformations.
22. Chymotrypsin can degrade neighboring molecules by catalyzing the hydrolysis of peptide bonds on the carboxyl side of
Phe, Tyr, and Trp. If the chymotrypsin is stored in a solution of
weak acid, His 57 would be protonated and would be unable to
accept a proton from Ser 195 to begin hydrolysis.
30. (a)
His 124
HN
CH
CH2
HN
HBr
CH
C
CH2
BrCH2COO
N
HN
N
N
CH2COO
[From Shapiro, R., Weremowicz, S., Riordan, J.F., and Vallee,
B., Proc. Natl. Acad. Sci. 84, 8783–8787 (1987).]
28. (a)
CH2
F
CH2
NO2
HF
NO2
O
NO2
14
| CHAPTER 6 Solutions
O
H
RNA Substrate
H
O
O
O
O
H
P
O
O
O
H
H
(b) His 124 most likely has a pK of less than 5.0 because
the His shown in the relay is unprotonated. It must be unprotonated in order to accept a proton from the water molecule. If His 124 had a higher pK value, it would be more
likely to be partially protonated at physiological pH and less
able to accept a proton.
(c) Alanine has an aliphatic side chain and is unable to accept a proton in the relay, as shown in part (a). [From Oda,
Y., Yoshida, M., and Kanaya, S., J. Biol. Chem. 268, 88–92
(1993).]
32. (a) His 12 has a pK value of 5.4 and His 119 has a pK value
of 6.4. His 12 is unprotonated and has the lower pK, while
His 119 is protonated and has the higher pK.
(b) The pH optimum is 6 because at this pH His 12 is
unprotonated (the pH is greater than the pK ) and His
119 is protonated (the pH is less than the pK ). At a pH
less than 6, His 12 would be protonated and would not
serve as a nucleophile to abstract the 29-hydroxyl hydrogen. At pH greater than 6, His 119 would be unprotonated and would not be able to donate a hydrogen to the
scissile bond.
(c) Ribonuclease does not hydrolyze DNA because DNA
contains deoxyribose and is missing the 29 hydroxyl group,
which serves as the attacking nucleophile for the phosphorous (once the His 12 has removed the hydrogen).
O
C
Asp 25
NO2
C
CH
C
N
34.
OH
HN
CH2
O
C
O
C
HN
26. His residues are often involved in proton transfer. A
carboxymethylated His would be unable to donate or accept
protons.
HN
CH
CH2
24. The Cys 278 is highly exposed and unusually reactive as
compared to other cysteines in creatine kinase. The Cys 278,
because of its high reactivity, is probably one of the catalytic
residues in the enzyme. The other cysteine residues are not
as reactive because they are not directly involved in catalysis
and/or because they are shielded in some way which prevents
them from reacting with NEM.
O
Asp 70
O
R2
H
O
O
H
C
O
O
NHR1
HO
C
Asp 25
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O
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R2
OH
C
HO
C
O
O
⫺
H
O
NHR1
Asp 25
O
C
O⫺
HO
H
Asp 25
40. The enzyme’s conformation must be flexible enough to allow substrates access to the active site, to stabilize the changing
electronic structure of the transition state, and to accommodate
the reaction products.
Asp 25⬘
R2
C
[From Shin, S., Yun, Y. S., Koo, H.M., Kim, Y.S., Choi, K.Y.,
and Oh, B.-H., J. Biol. Chem. 278, 24937–24943 (2003).]
C
O
HO
O
NHR1
42. (a)
C
O
Asp 25⬘
C
N⫹
36.
O
CH
HN
N
HN
CH
HN
CH2
H
Im
H
H
N
O
Zn2⫹
O
HN
NH
N⫹
C
C
Im
⫺
Im
Im
O
Zn2⫹
Im
Im
O
H 2O
H⫹
C
O
Im
H2O
Zn2⫹
Im ⫹ HCO⫺
3
Im
38.
CH2
O⫺
R2
H
C N
CH2
(CH2)4
HO
NH⫹
3
CH2
CH2
(CH2)4
NH⫹
3
R2
R1
O H O
H
C N R1
O⫺ First tetrahedral intermediate
O
H2N R1
Amino product
R2
CH2
CH2
(CH2)4
O
O⫺
NH⫹
3
C
H2O
H
O
R2
O
CH2
CH2
O
O
(CH2)4
NH⫹
3
C
Acyl–enzyme intermediate
O
H
CH2
O
R2
CH2
(CH2)4
CH2
CH2
(CH2)4
OH
NH⫹
3
O⫺
HO
NH⫹
3
C
OH
O
O⫺
Second tetrahedral intermediate
C
R2
⫹
O⫺
H⫹
⫺O
NO2
p-Nitrophenolate
H
O
⫹
NO2
O
C
CH2
O
NH
O
C
(b) The imidazole ring is already tethered to the nitrophenyl group, so the reaction is unimolecular rather than
bimolecular. In the bimolecular reaction, the reactants must
first encounter each other via diffusion in solution. The
unimolecular reaction proceeds faster because the reacting
groups are already in close proximity.
(c) Enzymes speed up reaction rates in part by proximity
and orientation effects. By binding to the enzyme, the reactants are in close proximity to the one another. The enzyme
also assists in binding the reactants in the proper orientation
so that the reaction can occur with less added free energy,
which results in a more rapid reaction.
44. A water molecule is able to enter the active site after the
first product, the C-terminal portion of the substrate, diffuses
away.
46. (a) The pocket that holds the P19 side chain is nonpolar,
fairly large, and able to hold side chains of varying sizes. The
Asp–Ala–Phe–Leu peptide is hydrolyzed fastest, followed
by Asp–Phe–Ala–Leu. Thus, the pocket can accommodate
small aliphatic side chains such as Ala as well as larger aromatic side chains such as Phe. A positively charged amino
acid does not fit well into this pocket, since Asp–Lys–Ala–
Leu is hydrolyzed more slowly than most of the other artificial peptides. The P29 pocket most likely accommodates a
large hydrophobic side chain, since the Asp–Ala–Phe–Leu
peptide was hydrolyzed faster than the peptides with Ala at
the P29 position. A substrate with a charged residue such as
Lys or Asp is hydrolyzed relatively slowly.
(b) Aspartame is a good candidate for hydrolysis by aspartyl aminopeptidase, since the enzyme catalyzes hydrolysis of
peptides on the carboxyl side of Asp residues (that is, Asp
is in the P1 position). The amino acid in the P19 position
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would be Phe, a large hydrophobic residue that should fit in
the P19 pocket, as suggested by the studies of the artificial
substrates.
(c)
O
⫹H
3N
CH
C
O⫺
CH2
O
⫹H N
3
CH
C
O
H
N
C
O
CH3
CH2
CH2
C
CH
C
aspartyl
aminopeptidase
O⫺
O
⫹H
O
O⫺
3N
CH
C
0.025 M
6.9 3 10 25 s
v 5 2360 M ? s21
[From Wolfenden, R., and Yuan, Y., J. Am. Chem. Soc. 13,
7548–7549 (2008).]
2d[P ]
d[S]
5
dt
dt
2 mol glucose
1.1 3 1023 M maltose
v5
3
s
1 mol maltose
v 5 2.2 3 1023 M ? s21
6. v 5 2
O
H2O
v52
OCH3
CH2
8. rate 5 k [sucrose]
The reaction is first-order overall.
Aspartame
(d) The enzyme likely consists of eight identical subunits
with a subunit molecular mass of 55 kD. The subunits are
associated with one another noncovalently, so the entire
enzyme complex has a molecular mass of 440 kD. [From
Wilk, S., Wilk, E., and Magnusson, R.P., J. Biol. Chem. 273,
15961–15970 (1998).]
48. Chymotrypsin activation is a cascade mechanism, since
chymotrypsinogen is activated by trypsin, which is in turn activated by enteropeptidase.
10. rate 5 k [sucrose]
rate 5 11.0 3 104 s21 2 10.050 M2
rate 5 5.0 3 102 M ? s21
12. (a)
(b)
[P]
Enteropeptidase
[S]
Time
(c)
Trypsinogen
(d)
Trypsin
v0
Chymotrypsinogen
Time
[ES]
Chymotrypsin
[E]
50. A trypsin inhibitor inactivates any trypsin that may have been
activated prematurely. This prevents activation of other pancreatic
zymogens, since trypsin is at the “top of the cascade.” Premature
activation of pancreatic zymogens results in destruction of pancreatic tissue.
52. A protease with extremely narrow substrate specificity (that
is, a protease with a single target) would pose no threat to nearby
proteins because these proteins would not be recognized as substrates for hydrolysis.
Chapter 7 Solutions
2. Enzyme activity is measured as an initial reaction velocity,
which is the velocity before much substrate has been depleted
and before much product has been generated. It is easier to measure the appearance of a small amount of product from a baseline of zero product than to measure the disappearance of a small
amount of substrate against a background of a high concentration of substrate.
d[S]
4. v 5 2
dt
16
| CHAPTER 7 Solutions
Time
(e)
v0
[S]
14. The enzyme concentration is comparable to the lowest substrate concentration and therefore does not meet the requirement that [E] ,, [S]. You could fix this problem by decreasing
the amount of enzyme used for each measurement.
16. v 0 5
v0 5
Vmax [S]
K M 1 [S]
165 mmol ? min 21 2 11.0 mM2
10.135 mM2 1 11.0 mM2
v 0 5 57 mmol ? min21
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18. The K M of enzyme A is about 2 mM and the K M for enzyme
B is about 5 mM.
(a) Enzyme A would generate product more rapidly when
[S] 5 1 mM.
(b) Enzyme B would generate product more rapidly when
[S] 5 10 mM.
Vmax 3 5 K M
5 KM 1 KM
Vmax 3 5 K M
v0 5
6 KM
v0 5 0.83 Vmax
20. v 0 5
When [S] 5 5 K M, the velocity is 83% of the maximum velocity.
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controlled, which means the reaction is catalyzed as rapidly
as the two reactants can encounter each other in solution.
Thus, enzymes B and C are diffusion-controlled but enzyme
A is not.
Enzyme
A
B
C
KM
k cat /K M
k cat
21
0.3 mM
1 nM
2 mM
5000 s
2 s21
850 s21
1.7 3 107 M21 ? s21
2 3 109 M21 ? s21
4.2 3 108 M21 ? s21
30. The Vmax can be calculated by taking the reciprocal of the
y intercept:
Vmax 5
1
y int
1
4.41 3 1024 mM21 ? h
v0 5
Vmax 3 20 K M
20 K M 1 K M
Vmax 5
v0 5
Vmax 3 20 K M
21 K M
Vmax 5 2.27 3 103 mM ? h21
v0 5 0.95 Vmax
When [S] 5 20 K M, the velocity is 95% of the maximum
velocity. Therefore, a fourfold increase in substrate concentration causes a smaller proportional increase in velocity (from
83% to 95%). Estimating V max from a plot of n0 versus [S]
is difficult because the substrate concentration must be quite
high in order to achieve a maximal velocity of close to 100%.
(And it is possible that these points on the hyperbolic curve
cannot be experimentally measured since at high concentration the substrate may not be soluble in the reaction medium.)
It is better to obtain V max by fitting experimental data to the
equation of a hyperbola or, alternatively, from a Lineweaver–
Burk plot.
22. kcat
Vmax
5
[E] T
4.77 3 1023 M
9.0 3 1026 M ? s21
5 530 s21
kcat 5
kcat
The k cat is the turnover number, which is the number of catalytic
cycles per unit time. Each molecule of the enzyme therefore undergoes 530 catalytic cycles per second.
24. The simultaneous collision of three molecules (E, A, and
B) is an unlikely event. It is much more likely that the enzyme
binds first one and then the other substrate. For example, the
first bimolecular reaction might be E 1 A z
y EA, and the second
would be EA 1 B z
y EAB.
26. The higher K M indicates that hexokinase has a lower affinity for fructose than for glucose. But once the substrate binds to
the enzyme, the fructose is converted to product more rapidly
than glucose.
28. The maximum rate at which two molecules can collide
with one another in solution is 108 to 109 M21 ? s21. Enzymes
with kcat/K M values in this range can be considered to be diffusion-
The K M can be determined by first calculating the x intercept and then by taking its reciprocal:
b
x int 5 2
m
4.41 3 1024 mM21 ? h
x int 5 2
0.26 mM21 ? h ? mM
x int 5 21.70 3 1023 mM21
1
KM 5 2
x int
1
KM 5 2
21.70 3 1023 mM21
K M 5 590 mM
32. (a) The wild-type enzyme has the lowest K M, indicating
that it has the highest affinity for aspartate. The aspartate substrate is not able to bind to the mutant enzymes
with as high an affinity. There is an 18-fold increase in
K M when the Arg 386 is mutated to a lysine and an 80fold increase in K M when Arg 292 is mutated to a lysine
(which is similar to the double mutant). The Arg 292
must play a critical role in binding the aspartate substrate
to the enzyme.
(b) Although both Arg and Lys have positive charges that
would attract the carboxylate group on the substrate,
there must be other interactions that are important in
addition to ionic interactions. For example, the ability
of Arg and Lys side chains to form hydrogen bonds differs. Also, steric considerations may play a role, since
the Lys side chain has a different shape than the Arg side
chain.
(c) Substitution of Arg with Lys greatly decreases the
catalytic efficiency of the mutant enzyme. Changing
only one arginine decreases the catalytic efficiency by
four orders of magnitude, but the catalytic efficiency decreases by six orders of magnitude when both arginines
are replaced.
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Enzyme
Wild-type Asp
AT (Arg 292
Arg 386)
Mutant Asp AT
(Lys 292 Arg
386)
Mutant Asp
AT(Arg 292
Lys 386)
Mutant Asp AT
(Lys 292 Lys
386)
K M Aspartate
kcat (s21)
(mM)
4
530
kcat/K M
(mM21 ? s21)
1.3 3 105
326
4.5
1.4 3 101
72
9.6
1.33 3 102
0.055
1.8 3 1021
300
(d) Changing the amino acid from Arg to Lys may cause
a three-dimensional conformational change that would
interfere with the ability of the catalytic apparatus to
function properly. If proton donation is important to the
mechanism, this would be affected as well, since the lysine
side chain donates a proton more readily than the resonance-stabilized guanidino group in the side chain of arginine. [From Vacca, R.A., Giannattasio, S., Graber, R.,
Sandmeier, E., Marra, E., and Christen, P., J. Biol. Chem.
272, 21932–21937 (1997).]
34. By irreversibly reacting with chymotrypsin’s active site,
DIPF would decrease [E]T. The apparent Vmax would decrease
since Vmax 5 kcat[E]T (Equation 7-23). K M would not be affected since the unmodified enzyme would bind substrate
normally.
36. (a) Indole is a competitive inhibitor of chymotrypsin because its structure resembles the side chain of tryptophan,
which fits into the specificity pocket of chymotrypsin. Thus,
indole and tryptophan side chains compete with each other
for binding to the active site.
(b) The Vmax is the same in the presence and absence of the
inhibitor since inhibition can be overcome at high substrate
concentrations. The K M increases because a higher concentration of substrate is needed to achieve half-maximal activity in the presence of an inhibitor.
38. Since competitive inhibitors compete with substrate for
binding to the active site of the enzyme, it is possible (especially if the concentration of the substrate increases in vivo)
that this inhibition can be overcome. There might also be
cellular complications if these competitive inhibitors, which
resemble substrates, accumulate. Noncompetitive and uncompetitive inhibitors do not resemble the substrate, and inhibition cannot be overcome by increasing the concentration of
substrate. If these inhibitors bind to their targets very tightly,
small amounts can be used to inhibit the enzyme effectively.
[From Westley, A.M., and Westley, J., J. Biol. Chem. 271,
5347–5352 (1996).]
18
| CHAPTER 7 Solutions
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40. The compound is a transition state analog (it mimics the
planar transition state of the reaction) and therefore acts as a
competitive inhibitor.
42. (a) Zanamivir, with a lower K I (indicating tighter binding
of the inhibitor to the enzyme) would work better.
(b) Vmax is about the same, but K M increases, so the mutant
enzyme probably binds the substrate more poorly than the
wild-type enzyme but does not exhibit any decrease in turnover number (kcat, reflected in V max).
(c) The mutation increases the K I for oseltamivir by |265
times but increases the K I for zanamivir by only |2 times.
Therefore, zanamivir would be a better inhibitor of neuraminidase in the mutant virus. [From Collins, P.J., Haire,
L.F., Lin, Y.P., Liu, J., Russell, R.J., Walker, P.A., Skehel,
J.J., Martin, S.R., Hay, A.J., and Gamblin, S.J., Nature 453,
1258–1261 (2008).]
44. The Vmax in the absence of inhibitor can be calculated by
taking the reciprocal of the y intercept:
Vmax 5
1
y int
Vmax 5
1
1.51 1OD21 ? min2
Vmax 5 0.66 OD ? min21
The Vmax in the presence of inhibitor can be calculated
similarly:
Vmax 5
1
y int
Vmax 5
1
4.27 1OD21 ? min2
Vmax 5 0.23 OD ? min21
The KM value in the absence of inhibitor and can be determined by first calculating the x intercept and then by taking its
reciprocal:
x int 5 2
b
m
x int 5 2
1.51 OD21 ? min
1.52 min ? OD21 ? mM
x int 5 20.99 mM21
1
KM 5 2
x int
1
KM 5 2
20.99 mM21
K M 5 1.0 mM
The K M value in the presence of inhibitor can be similarly
determined:
b
x int 5 2
m
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x int 5 2
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4.27 OD21 ? min
1.58 min ? OD21 ? mM
10 mM
KI
0.68 5 10mM/K I
K I 5 15 mM
1.68 5 1 1
x int 5 22.70 mM21
1
KM 5 2
x int
1
KM 5 2
22.70 mM21
K M 5 0.37 mM
[From Paulus, C., Hellebrand, S., Tessmer, U., Wolf, H.,
Kräusslich, H.-G., and Wagner, R., J. Biol. Chem. 274, 21539–
21543 (1999).]
Dodecyl gallate is an uncompetitive inhibitor. In the presence of the inhibitor, the Vmax and the K M values decreased to
a similar extent. The slopes of the lines in the Lineweaver–Burk
plot are nearly the same. [From Kubo, I., Chen, Q.-X., and
Nihei, K.-I., Food Chemistry 81, 241–247 (2003).]
46. (a) Lineweaver–Burk plots are shown below. The K M is calculated from the x intercept, the Vmax from the y intercept.
48. (a) The Lineweaver–Burk plot is shown. The K M is calculated from the x intercept, the Vmax from the y intercept.
21
y intercept, (mM/min)
Vmax (mM/min)
x intercept (mM)21
K M (mM)
Without inhibitor With inhibitor
0.704
1.90
1.42
0.52
–0.949
–2.54
1.05
0.39
Inhibition of HIV-1 protease by p6*
5.0
0.4
⫺0.02
1/v0 (mM −1 . min)
1/v0 (min/nmol)
y ⫽ 3.4337x ⫹ 0.038
0.3
0.2
y ⫽ 1.8985x ⫹ 0.0349
0.1
0
0.02
0.04
1/[S]
0.06
0.08
Without p6*
K M (mM)
y intercept (min ?
nmol21)
Vmax (nmol ? min21)
3.0
2.0
0.1
y ⫽ 0.742x ⫹ 0.704
1.0
20.0185
54
⫺3.0 ⫺2.5 ⫺2.0 ⫺1.5 ⫺1.0 ⫺0.5 0.0 0.5
1/[S] (mM⫺1)
With p6*
20.011
89
0.035
28.5
0.0385
26.0
(b) The inhibitor is a competitive inhibitor. The Vmax is the
same in the presence and absence of the inhibitor (within
experimental error), but the K M has increased nearly twofold, indicating that the p6* is competing with the substrate
for binding to the active site of the enzyme.
(c)
y ⫽ 0.751x ⫹ 1.905
(␮M⫺1)
Without inhibitor
With inhibitor
x intercept (mM21)
4.0
21/aK M
1
5
a
21/K M
120.011 mM21/20.0185 mM21 2 5 0.595
a 5 1.68
a511
1I2
KI
No inhibitor
1.0
1.5
2.0
2.5
3.0
With homoarginine
(b) Homoarginine is an uncompetitive inhibitor. The slopes
of the lines in the Lineweaver–Burk plot are nearly identical. A proportional decrease in Vmax and K M occurs in the
presence of the inhibitor.
(c) Because homoarginine is an uncompetitive inhibitor,
it does not bind to the active site of the alkaline phosphatase enzyme but to another site that interferes with
the activity of the enzyme in some way. The intestinal
alkaline phosphatase catalyzes the same reaction as the
bone alkaline phosphatase, so the active sites of the two
enzymes are likely to be similar, but the structures of the
enzymes may be sufficiently different that the intestinal
enzyme lacks the binding site for homoarginine. [From
Lin, C.-W., and Fishman, W.H., J. Biol. Chem. 247,
3082–3097 (1972).]
50. The formation of a disulfide bond under oxidizing conditions, or its cleavage under reducing conditions, could act as an
allosteric signal by altering the conformation of the enzyme in a
way that affects the groups at the active site.
CHAPTER 7 Solutions | 19
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Chapter 8 Solutions
2.
H3C
(CH2)4
CH
CH
CH2
CH
CH
(CH2)4
5, 11, 14
Sciadonate (20:3
CH
CH
(CH2)3
COO
)
[From Sayanova, O., Haslam, R., Venegas Caleron, M., and
Napier, J. A., Plant Physiology 144, 455–467 (2007).]
10. Phosphatidylcholine and phosphatidylethanolamine are
neutral. Phosphatidylglycerol and phosphatidylserine are negatively charged.
12.
(a)
O
O
4.
O
CH2
HO
C
O
C
(CH2)7
H
CH3
C
C
O
CH2
(CH2)7
CH3
CH2
CH
H2C
OH
O
COO
O
R
C
OH
CH
NH
3
O
H
CH2
P
O H
C
O
R
[From Chao-Mei Y., Curtis, J.M., Wright, J.L.C., Ayer, S.W.,
and Fathi-Afshar, Z.R., Can. J. Chem. 74, 730–735 (1996).]
(b)
H2C
R1 R2
6.
O
C
C
O
O
H2C
O
O
HO
H
CH
CH2 O
P
C
C
R
R
CH3
O
(CH2)2
O
N
CH3 H
CH3
O
O
O
P
O
O
P
O
O
H
H HO
OH H
O
CH2 HO
CH
O
O
C
C
R
R
CH2
CH
CH2
OH
OH
O
H
OH
H
OH
OH
OH
H2C
O
H2C
H
8.
O
CH2
O
O
O
CH
(c)
CH
OH
CH
14. Vitamins A, D, and K are isoprenoids and are nonpolar.
These dietary vitamins are soluble in the synthetic lipid Olestra®
and pass out of the intestinal tract along with the Olestra® without being absorbed. Adding these vitamins to the product helps
saturate the synthetic lipid with vitamins so that dietary vitamins
are not excreted.
16. A glycerophospholipid with two saturated acyl chains has
a cylindrical shape, whereas a glycerophospholipid with two
unsaturated, kinked acyl chains would be more cone-shaped:
NH
3 CH
CH
(CH2)2
CH
C
CH3
CH
CH
(CH2)6
CH3
[From Ohasi, Y., Tanaka, T., Akashi, S., Morimoto, S., Kishimoto, Y., and Nagai, Y., J. Lipid Res. 41, 1118–1124 (2000).]
20
| CHAPTER 8 Solutions
Saturated acyl chains
Unsaturated acyl chains
18. Phospholipase A1 catalyzes the hydrolysis of one of the acyl
chains on a phospholipid. The resulting product, a lysophospholipid
(see Problem 12a) has a cone-shaped structure (see Fig. 8-4). Lysophospholipids do not assemble to form bilayers as the cylindricalshaped glycerophospholipids do. The conversion of a significant
portion of glycerophospholipids to lysophospholipids results in
the destruction of the red blood cell membrane.
20. B . A . C. Trans-oleate has a melting point similar to
that of stearate (18:0) because the trans double bond does not
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produce a kink in the molecule. Its geometry more closely resembles that of a single bond.
22. Peanut oil has a higher melting point because the fatty acids
that compose the monounsaturated triacylglycerols have a higher
melting point than the more highly unsaturated fatty acids of
the vegetable oil. Each double bond introduces a “kink” in the
molecule, which means that the fatty acids don’t pack together as
well. The number of double bonds corresponds to the number of
“bends” in the acyl chain. Fatty acids that do not pack together
well have fewer London dispersion forces among the chains, and
less heat energy is required to disrupt the forces and melt the
solid. Therefore, the vegetable oil, with a higher percentage of
polyunsaturated triacylglycerols, has a lower melting point and
does not freeze.
24. When phytanic acid is incorporated into membrane phospholipids, the resulting membrane is more fluid. The presence of
the methyl groups on the phytanic acid results in an acyl chain
that has a decreased ability to interact with neighboring acyl
chains. This decreases the number of van der Waals interactions
and decreases the melting point, which increases membrane fluidity. [From van den Brink, D.M., van Miert, J.N.I., Dacremont,
G., Rontani, J.-F., and Wanders, R.J.A., J. Biol. Chem. 280,
26838–26844 (2005).]
26. Increasing the temperature would make the membrane
more fluid. To maintain constant fluidity, the bacteria synthesize fatty acids with more carbons and with fewer double
bonds.
28. In a bilayer, one end of each lipid acyl chain is fixed by its
attachment to a head group. The methylene groups closer to the
head group have the least conformational freedom, whereas the
methylene groups farthest from the head group, near the bilayer
center, have the greatest freedom.
30. The cold temperatures might induce the activation of
desaturase enzymes in the plant that convert the 18:0, 18:1, and
18:2 fatty acids to 18:3 fatty acids. Of these four fatty acids,
18:3 is the most unsaturated and has the lowest melting point.
Membranes composed of phospholipids containing unsaturated
fatty acids maintain their fluidity in cold temperatures. [From
Shi, Y., An, L., Zhang, M., Huang, C., Zhang, H., and Xu, S.,
Protoplasma 232, 173–181 (2008).]
32. The flippase enzyme has a strong preference for PS,
which is consistent with the observation that PS is located
exclusively in the cytosolic-facing leaflet of the membrane.
Translocation of PE also explains the predominance of PE
in the cytosolic leaflet. PC and SM, both choline-containing lipids, are not translocated and remain in the extracellular leaflet of the membrane. The flippase has a preference
for amino-containing phospholipid head groups and eschews
choline-containing phospholipids. The flippase requires ATP,
probably because translocation occurs against a concentration
gradient, and magnesium ions. A cysteine amino acid side
chain must be essential for translocation activity since chemical modification of this side chain abolishes translocation activity. [From Daleke, D.L., and Heustis, W.H., Biochemistry
24, 5406–5416 (1985).]
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34. (a) A or D
(b) A or C
(c) D
(d) D
(e) B
(f ) B or E
36. Cytochrome c is a peripheral membrane protein that is
only loosely associated with the membrane and can therefore be
removed by gentle means such as a salt solution. Cytochrome
oxidase is an integral membrane protein that completely spans
the bilayer and has large hydrophobic portions where the protein
spans the bilayer. Thus, it is difficult to remove unless nonpolar
organic solvents or amphipathic detergents are used to dissociate
it from the membrane.
Peripheral
Integral
38. (a) A fully hydrogen-bonded b barrel can form only if the
number of strands is even. A b sheet with an odd number of
strands could not close up on itself to form a barrel.
(b) The strands are antiparallel because adjacent strands can
be easily linked by loops on the solvent-exposed portions of
the protein.
(c) A b barrel could contain some parallel b strands, but
these could not be consecutive. A b barrel with consecutive parallel strands could occur only if the strands were
linked by additional membrane-spanning segments (such as
a transmembrane a helix or a structure that passed through
the center of the barrel).
40. Melittin is more conformationally restricted in the membrane when it is associated with glycerophospholipids containing
more saturated fatty acids (like oleate, 18:1) and less restricted
when associated with lipids containing more unsaturated fatty
acids (like arachidonate, 20:4). The presence of double bonds
in the acyl chains produces “bends” that prevent the acyl chains
from packing closely together. The unsaturated fatty acids thus
provide a less conformationally restricted environment for the
melittin peptide. In contrast, when there are fewer double bonds,
such as in oleate, the fatty acyl chains pack more closely together
and interact via van der Waals forces. This environment restricts
the conformation of melittin.
42. The sandwich model assumes that all membranes are the
same. However, it has been noted that different membranes have
a variety of functions. Multiple functions would not be possible
with the same structure. The model also does not explain how
membrane transport could occur, since it would be difficult for
transported molecules to get past the protein “caps” on either
side of the membrane.
CHAPTER 8 Solutions | 21
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44. The bleached area “recovers” its fluorescence as fluorophorelabeled molecules diffuse out of the small area and unbleached
fluorophore-labeled molecules diffuse in. These experiments are
useful for measuring diffusion rates of the target molecules.
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(b)
DG 5 RT ln
[H 1 ] in
1 ZFDc
[H 1 ] out
1026.88
1027.78
21
21
1 112 196485 J ? V ? mol 2 120.052 V2
DG 5 18.3145 J ? K21 ?mol21 2 137 1 273 K2 ln
Chapter 9 Solutions
2.
[Na 1 ]in
Dc 5 0.058 log
[Na 1 ]out
[Na 1 ]in
10.050 5 0.058 log
[Na 1 ]out
1
[Na ]in
0.86 5 log
[Na 1 ]out
[Na 1 ]in
100.86 5
[Na 1 ]out
[Na 1 ]in
7.3
5
1
[Na 1 ]out
When a nerve cell is depolarized, sodium ions enter the cell;
therefore the [Na1]in /[Na1]out ratio is more than 100-fold greater
in the depolarized cell than in the resting cell.
4. Use Equation 9-4 and let Z 5 2 and T 5 310 K :
(a)
DG 5 RT ln
[Ca21 ]in
1 ZFDc
[Ca21 ]out
1027
1023
21
21
1 122 196,485 J ? V ? mol 2 120.05 V2
DG 5 223,700 J ? mol2129600 J ? mol21
DG 5 233,300 J ? mol21
DG 5 18.3145 J ? K21 ? mol21 2 1310 K2ln
The negative value of DG indicates a thermodynamically
favorable process.
(b)
DG 5 RT ln
[Ca21 ]in
1 ZFDc
[Ca21 ]out
10 27
10 23
21
21
1 122 196,485 J ? V ? mol 2 110.05 V2
DG 5 18.3145 J ? K 21 ? mol 21 2 1310 K2 ln
DG 5 223,700 J ? mol21 1 9600 J ? mol21
DG 5 214,100 J ? mol21
The negative value of DG indicates a thermodynamically unfavorable process, but not as unfavorable as in part (a).
6. (a)
Dc 5 0.058 log
Dc 5 0.058 log
[H 1 ]in
[H 1 ]out
11027.78 2
11026.88 2
Dc 5 20.052 V 5 252 mV
22
| CHAPTER 9 Solutions
DG 5 15344 J ? mol21 1 25017 J ? mol21
DG 5 327 J ? mol21
[From Porcelli, A.M., Ghelli, A., Zanna, C., Pinton, P.,
Rizzuto, R., and Rugolo, M., Biochem. Biophys. Res. Commun.
326, 799–804 (2005).]
8. The less polar a substance, the faster it can diffuse through
the lipid bilayer. From slowest to fastest: C, A, B.
10. The D-E-K-A sequence already contains two negatively
charged side chains, so the lysine (K, which is positively charged)
and the alanine (A, which is neutral) can be mutated to negatively charged amino acid side chains, either Asp (D) or Glu (E).
[From Miedema, H., Meter-Arkema, A., Wierenga, J., Tang,
J., Eisenberg, B., Nonner, W., Hektor, H., Gillespie, D., and
Meijberg, W., Biophysical Journal 87, 3137–3147 (2004).]
12. At neutral pH, the Asp and Glu side chains are unprotonated and can participate in ion pairing that holds the protein in
its closed conformation. In the presence of acid (low pH, high
[H1]), the Asp and Glu side chains become protonated. This
disrupts ion pairing, and the conformation of the protein shifts
to an open state.
14. (a) Rhcg appears to encode an ammonia channel, since its
absence decreases the amount of NH3 that crosses the cell
membrane.
(b) Because the large-scale transport of water (a small polar molecule) requires aquaporin, it is not surprising that
the transport of ammonia (also a small nonpolar molecule)
requires a transport protein, especially in the kidneys, which
are responsible for ammonia excretion. [From Biver, S.,
Belge, H., Bourgeois, S., Van Vooren, P., Nowik, M., Scohy,
S., Houillier, P., Szpirer, J., Szpirer, C., Wagner, C.A., Devuyst,
O., and Marini, A.M., Nature 456, 339–343 (2008).]
16. (a) Because this sequence is so similar to the KcsA and NaK
sequences, it probably comes from a channel that is specific
for K1 or for Na1 and K1.
(b) This sequence is most similar to the Ca21 II channel
sequence and so probably comes from a Ca21 channel.
18. Intracellular exposure of the glucose transporter to trypsin
indicates that there is at least one cytosolic domain of the transport protein that is essential for glucose transport. Hydrolysis of
one or more peptide bonds in this domain(s) abolishes glucose
transport. But extracellular exposure of the ghost transporter to
trypsin has no effect, so there is no trypsin-sensitive extracellular domain that is essential for transport. This experiment also
shows that the glucose transporter is asymmetrically arranged in
the erythrocyte membrane.
20. As the glutamate (charge 21) enters the cell, 4 positive charges
also enter (3 Na1, 1 H1) for a total of 3 positive charges. Since
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1 K1 exits the cell at the same time, a total of 2 positive charges
are added to the cell for each glutamate transported inside.
22. Treatment with ouabain “freezes” the Na,K-ATPase in the
phosphorylated conformation, and the reaction cycle cannot be
completed. As a result, the pump is nonfunctional and the concentration of sodium ions inside the cell increases. Water enters
the cell along with the sodium ions, and the cells will eventually
lyse due to the increased osmotic pressure.
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28. (a) V max is calculated from the y intercept.
O
C
⫹
O
CA II
C
O
Water
Carbon dioxide
⫹
K M is calculated from the x intercept (without inhibitor):
x int 5 2b /m
0.64 g ? min ? mmol21
x int 5
5.4 g ? min ? mmol21 ? mM
H⫹
Bicarbonate
x int 5 20.12 mM21
1
KM 5 2
x int
1
KM 5 2
20.12 mM21
K M 5 8.3 mM
(b) Bone-resorbing
Na+
K+
H+
1
0.65 g ? min ? mmol21
OH
HCO⫺
3
compartment
pH = 5.5
1
0.64 g ? min ? mmol21
I
Vmax
5 1.5 mmol ? g21 ? min21
O Carbonic acid
H2CO3
Vmax 5
I
Vmax
5
OH
H
1
y int
Vmax 5 1.6 mmol ? g21 ? min21
24. (a)
H
Vmax 5
Na+
CA II
H2O + CO2
H2CO3
H2CO3
H+ + HCO3−
app
K M is calculated from the slope (with inhibitor)
x int 5 2b /m
CI−
pH = 7
HCO3−
Hydrogen ions are produced from carbonic acid, which is
produced from water and carbon dioxide via the carbonic
anhydrase reaction. The hydrogen ions exit the cell via
the Na1/H1 exchanger. These hydrogen ions acidify the
bone-resorbing compartment as sodium ions enter the
cell. The sodium ions leave the cell via the Na,K-ATPase,
while potassium ions enter the cell. Bicarbonate, the
other product of the dissociation of carbonic acid, leaves
the cell via the Cl–/HCO 3– exchanger as chloride enters
the cell.
1
26.
KM 5 2
x int
1
5 4 mM
20.25 mM
1
5
y int
KM 5 2
V max
1
0.030 mg ? h ? nmol21
5 33 nmol ? mg21 ? h21
V max 5
V max
[From Wakisaka, M., Yoshinari, M., Yamamoto, M., Nakamura,
S., Asano, T., Himeno, T., Ichikawa, K., Doi, Y., and Fujishima,
M., Biochim. Biophys. Acta 1362, 87–96 (1997).]
x int 5
0.65 g ? min ? mmol21
13.8 g ? min ? mmol21 ? mM
x int 5 20.047 mM21
1
K app
M 5 2
x int
1
K app
M 5 2
20.047 mM21
app
K M 5 21.2 mM
(b) Phlorizin is a competitive inhibitor. The Vmax has remained nearly unchanged while the K M increased nearly
threefold in the presence of the inhibitor. Phlorizin competes with glucose for binding to the transporter. [From
Betz, A.L., Drewes, L.R., and Gilboe, D.D., Biochim. Biophys. Acta 406, 505–515 (1975).]
30. The first vinblastine binds to a low-affinity binding site,
the second vinblastine to a high-affinity site. This indicates that
when vinblastine binds to LmrA, it does so cooperatively. A conformational change occurs that allows the second molecule of
vinblastine to bind with greater affinity than the first.
32. Ammonium ions exiting the cell could be exchanged for
entering Na1 ions. The free energy for NH41 transport would be
provided directly by the movement of Na1 ions down their gradient, and indirectly by ATP, which is used by the Na,K-ATPase
to establish the Na1 gradient.
34. In the presynaptic cell, when an action potential reaches the
axon terminus, voltage-gated Ca21 channels open, flooding the
CHAPTER 9 Solutions | 23
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cell with Ca21 ions and triggering exocytosis of the acetylcholinecontaining synaptic vesicles. Acetylcholine diffuses across the
synaptic cleft and binds to receptors on the postsynaptic cell,
triggering a series of events that results in muscle contraction (see
Fig. 9-16). If antibodies block the Ca21 channels, acetylcholine
is not released from the presynaptic cell, receptors on the postsynaptic cell are not activated, and muscle contraction does not
occur. Thus, patients with this autoimmune disorder suffer from
muscle weakness.
36. If the serine in the active site of acetylcholinesterase is
covalently modified, the enzyme will not be able to function.
Acetylcholine will build up in the synaptic cleft and will not
be hydrolyzed. The postsynaptic cell remains polarized and does
not return to its resting state, so it cannot receive the next nerve
impulse from the presynaptic cell.
38. Botulinum toxin destroys SNAREs, which are required for
the fusion of synaptic vesicles with the neuronal plasma membrane. This interrupts communication between facial nerves and
muscles. The result is paralysis of the muscles whose contraction
accentuates wrinkles.
40. Diacylglycerol is a lipid without a phosphate-derivative
head group. Because it consists mostly of lipid “tails,” it could
promote the inward curvature of the bilayer, which occurs during membrane fusion (see Fig. 9-19).
2. Prednisone is a steroid, a hydrophobic molecule that can diffuse
through the lipid bilayer without the need for a cell-surface receptor.
The receptor for prednisone is located either in the cytosol or in the
nucleus. Upon binding to its receptor, the ligand–receptor complex
binds to DNA and induces transcription of specific genes.
(b)
24
[RL]
[ L]
5
[R]T
K d 1 [L]
Kd
[RL]
5
5
Kd
[R]T
Kd 1
5
0.2 K d
[RL]
5
[R]T
K d 1 0.2 K d
0.2 K d
[RL]
5
[R]T
1.2 K d
[RL]
[R]T
[RL]
[R]T
[RL]
[R]T
[RL]
[R]T
[RL]
[R]T
(c)
[ RL]
[L]
5
[R]T
K d 1 [L]
5 Kd
[RL]
5
[R]T
Kd 1 5 Kd
5 Kd
[RL]
5
[R]T
6 Kd
[ RL]
5 0.83
[R]T
6. If the number of receptors decreases to 150, a greater proportion
of these receptors must have bound ligand, since occupation of
100 receptors is required for a maximal response (see Problem 5).
A 20-fold increase in the concentration of ligand is required in
order to achieve a maximal response.
[ RL]
[L]
5
[ RT ]
[L] 1 K d
[L]
100
5
150
[L] 1 1 .0 3 10 210 M
0 .671 [L] 1 1 .0 3 10 210 M2 5 [L]
0 .67[ L] 1 6 .7 3 10 211 M 5 [L]
6 .7 3 10 211 M 5 0 .33[L]
Chapter 10 Solutions
4. (a)
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5 0.17
[ L]
K d 1 [L]
Kd
5
Kd 1 Kd
Kd
5
2 Kd
5
5 0.50
| CHAPTER 10 Solutions
[L] 5 2 .0 3 10 210 M
8. First the cells would be lysed and insoluble components
removed by centrifugation. The membranes and the associated
membrane proteins can be solubilized by adding a detergent.
An affinity column is constructed by covalently attaching a
specific ligand to the chromatography matrix. The solubilized
membranes are then loaded onto the column. Cellular proteins
that do not bind to the ligand will pass through the column.
The receptor proteins are then eluted from the column using a
concentrated salt solution to disrupt the intermolecular interactions between the receptor and the ligand.
10. When glucagon and epinephrine bind to their respective G
protein–coupled receptors, the result is the same—an enzyme
is activated that results in the synthesis of the second messenger cAMP. Since the binding of both ligands to their receptors
results in the activation of the same second messenger, the same
signal transduction pathway is activated and the same cellular
response is observed.
12. The strategy shown in the figure results in amplification
of the signal, since a single molecule of A can activate many
molecules of B, which can then go on to activate even more
molecules of C, and finally D. Amplification of a signal is an
advantage to the cell because a small signal (i.e., a low concentration of A) can result in a large response (a high concentration
of D). The strategy shown allows cells to respond to changing
conditions by dramatically altering the activities of intracellular
components.
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14. Both hormones lack tyrosine’s carboxylate group and have
hydroxyl groups attached to the ring and to the b carbon. In
epinephrine, the amino nitrogen bears a methyl group.
16. The inhibition of the intrinsic GTPase activity results in
a continuously active G protein. This increases the activity of
adenylate cyclase, which results in an increase in the concentration of intracellular cAMP. In intestinal cells, the increase
cAMP concentration leads to the loss of water and electrolytes
from the cells and results in diarrhea that can potentially be
fatal.
18. GTPgS can bind to a G protein, but since it cannot be
hydrolyzed, the G protein is in a persistently active state. If
GTPgS binds to a stimulatory G protein, then adenylate cyclase
is continually active, which has the effect of increasing cellular cAMP concentration. If GTPgS binds to an inhibitory G
protein, adenylate cyclase is continually inhibited, and cellular
cAMP concentration decreases.
20.
N
CH2
N
PO32
30. NO synthase is activated when calcium ions bind to
calmodulin (see Solution 27). Clotrimazole is a calmodulin antagonist and prevents NO synthase from binding to calmodulin. In the absence of active NO synthase, cyclic GMP is not
produced and protein kinase G is not activated. Cellular targets
of protein kinase G are not phosphorylated in the presence of
clotrimazole.
22.
OH
HC
O
R
C
HN
CH
CH
(CH2)12
CH3
O
CH
CH2
O
P
Sphingomyelin
CH3
O
CH2
CH2
O
N
CH3
CH3
32. Inhibition of cGMP phosphodiesterase, the enzyme that
hydrolyzes cGMP, increases the intracellular cGMP concentration. The activity of protein kinase G is increased as a result, as is
the phosphorylation of the enzyme’s targets, which are proteins
involved in smooth muscle cell relaxation. The overall result is
increased blood flow.
sphingomyelinase
OH
HC
O
R
C
HN
CH
CH
CH
CH2
(CH2)12
CH3
OH
O
Ceramide
O
P
CH3
O
CH2
CH2
O
described in Problem 25, activation of protein kinase B promotes cell growth, so a highly active protein kinase B could
account for the uncontrolled cell growth observed in cancer
cells.
28. (a) Acetylcholine in the synaptic cleft is rapidly hydrolyzed
by acetylcholinesterase (see Section 9-4).
(b) Various calcium pumps, using the energy of ATP hydrolysis, pump calcium ions out of the cytosol. The Ca21
concentration in the cell drops, the ion channels close,
and Ca21 ions dissociate from calmodulin, changing its
conformation and rendering it unable to bind to NO synthase. Without bound calmodulin, the NO synthase is
inactive.
(c) The signaling molecule NO rapidly decomposes. In
its absence, guanylate cyclase is not activated. Any cGMP
still present is hydrolyzed to GMP by cGMP phosphodiesterase.
(d) In the absence of cGMP, protein kinase G reassociates with its regulatory subunits and is inactivated. Any
phosphorylated proteins are acted upon by phosphatase enzymes, which remove the phosphate groups by
hydrolysis.
N
CH3
CH3
Phosphocholine
24. If calcineurin is bound to cyclosporine A, then calcineurin will be inactive as a phosphatase and will be unable
to catalyze the removal of the phosphate group from NFAT.
The NFAT remains in the cytosol, and the specific genes required for T cell activation are not expressed. The T cell, an
important component of the immune response, is not activated, which accounts for the immunosuppressive properties
of cyclosporine A.
26. Yes, mutations in which the gene coding for PTEN is
either nonfunctional or absent are commonly found in cancer
cells. In the absence of PTEN, inositol trisphosphate is not
dephosphorylated, and protein kinase B is highly active. As
34. In the presence of Ras with this particular mutation, the
signaling pathway would be constitutively active. Ras is active
when bound to GTP and is inactivated when GTP is hydrolyzed
to GDP. If hydrolysis of GTP cannot occur, Ras cannot be inactivated and the signaling pathway cannot be turned off. The
cell is stimulated to grow and proliferate even in the absence of
growth-signaling ligand.
36. Overexpression of IRS-1 would stimulate the activity of the
signaling pathway and would lead to enhanced activation of protein kinases B and C. The activation of protein kinase B would
result in the activation of glycogen synthase, so the cultured cells
would be expected to show an increase in glycogen synthesis.
Enhanced activation of protein kinase C would stimulate translocation of glucose transporters to the plasma membrane; thus,
an increase in glucose import would be observed in the cultured
cells.
38. A ligand could bind to a G protein–coupled receptor and
subsequently activate phospholipase C. This enzyme catalyzes
the cleavage of phosphatidylinositol bisphosphate to diacylglycerol
CHAPTER 10 Solutions | 25
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and inositol trisphosphate. Diacylglycerol activates protein kinase C, which then activates the MAP kinase cascade, as shown
in Problem 37. This is an example of cross-talk, in which signaling pathways share intracellular components.
40. (a) In order to treat cancer in these tissues, membranepermeable drugs could be developed that would interact
with intracellular steroid receptors to prevent binding by
the steroid ligand.
(b) Because the activation of Ras is a common feature
of many transformed cells (see Box 10-B), it’s possible
that some steroids use a second signaling mechanism
that involves activation of Ras, either through binding
to G protein–coupled receptors or to receptor tyrosine
kinases.
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10.
46. Aspirin inhibits cyclooxygenase (see Box 10-C) by acetylating an essential Ser residue on the enzyme. In the presence
of aspirin, thromboxanes are not synthesized. Thromboxanes
promote platelet aggregation and vasoconstriction, which could
promote clot formation and high blood pressure and could lead
to a heart attack.
HO
26
H
O
HOCH2
H
OH
HO
H
OH
CH2OH
OH
H
H
␤-D-Fructose
(b) A six-membered ring results.
H
H
O
H
H
H
HO
HO
O
H
CH2OH
OH
H
H
HO
H
OH
HO
OH
OH
CH2OH
H
␤-D-Fructose
␣-D-Fructose
14. The b anomer is more stable because all of the bulky substituents (—OH and —CH2OH groups) are in the equatorial
position on the chair conformation of glucose. The a anomer
has one hydroxyl group in the axial position, so the equilibrium
position favors formation of the b anomer and the concentrations of the anomers are not equal.
16.
O
C
O
⫹H N
3
H
OH
CH
C
H
O⫺
H
⫹
H
OH
CH2
OH⫺
(CH2)4
H
C
OH
OH
CH2OH
18. There are several possibilities; one is shown here.
CH2OH
HCOH
H
O
OH
H
H
OH
H
OH
CH2OH
CH2OH
H
OH
O
H
OH
H
O
H
O
H
OH
H
OH
OH
H
H
HO
H
H
COO⫺
NH⫹
3
H
H
H
NH3⫹
CH2OH
N⫹
OH
HO
CH2
CH2
C
H
CH2
OH
20.
| CHAPTER 11 Solutions
OH
HO
OH
8. The b anomer is more stable because most of the bulky
hydroxyl substituents are in the equatorial position.
␤-Mannose
OH
␣-D-Fructose
H
H
H
␤-D-Ribose
CH2OH
H
6. Fructose and galactose are isomers of glucose.
HO
O
H
2. Coenzyme A, NAD, and FAD all contain ribose residues.
CH2OH O
HO
H
HO
H
H
OH
HOCH2
H
H
OH
HO
OH
OH
H
H
12. (a) A five-membered ring results.
HO
4. (a) D-Psicose and D-sorbose are epimers.
(b) D-Sorbose and D-fructose are structural isomers.
(c) D-Fructose and L-fructose are enantiomers.
(d) D-Ribose and D-ribulose are structural isomers.
H
O
H
H
H
H
␣-D-Ribose
H
Chapter 11 Solutions
H
O
H
42. Protein kinase B (Akt) is anti-apoptotic (see Problem 25).
If S1P activates protein kinase B, then the cell will not undergo
programmed cell death but will grow and proliferate. This explains S1P’s ability to promote cell survival, as shown in the diagram in Problem 41.
44. Both ceramide kinase and sphingosine kinase are potential drug targets, since these enzymes catalyze the synthesis of
C1P and S1P, respectively, and both of these products promote
cell survival. Inhibiting the enzymes that catalyze production of
these pro-survival signaling molecules may inhibit the growth
of cancer cells. But caution should be exercised, since inhibitors of COX-2, such as Vioxx, were determined to be unsuitable
drugs because of unexpected side effects.
H
OH
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Cellobiose is a reducing sugar. The anomeric carbon of the
glucose on the right side is free to reverse the cyclization reaction to re-form the aldehyde functional group, which can be
reduced.
structure. Consequently, iodine does not form a complex with
the cellulose and a blue color is not produced.
34.
H
22. Trehalase digestion produces glucose, which exists in solution as a mixture of the a and b anomers.
H
HO
CH2OH
O
H
OH H
H
H
H
OH
HO
OH
CH2OH
O
H
OH H
H
H
HO
H
OH
HO
H
H
HO
OH
OH
H
H
OH
H
O
H
OH
H
H
OH
O
H
36.
CH2OH
OH
H
O
OH
H
OH
H
H
OH
OH
26. Celery is mainly cellulose and water, neither of which
provides nutritive calories. Humans do not have b-glucosidase
enzymes and cannot hydrolyze the b-glycosidic bonds linking
the glucose resides in cellulose. Since cellulose is not digested,
the body does not spend any energy to further process it. Foods
like celery contribute roughage, or fiber, to the diet, but these
foods neither provide nor cost the body much in the way of
energy.
28. The fungus must have contained an a-galactosidase enzyme
capable of digesting the a(1S6) linkage and a sucrase-like enzyme that could hydrolyze the (1S2) glycosidic bond formed
between the a anomer of glucose and the b anomer of fructose.
[From Feng, S., Saw, C.L., Lee, Y.K., and Huang, D., J. Agric.
Food Chem. 56, 10078–10084 (2008).]
30. There is one reducing end.
32. Potato starch consists of a mixture of amylose and amylopectin. The iodine can fit into the center of the helical structure
of the amylose and produce a blue color. Apples, on the other
hand, contain mostly cellulose. Cellulose consists of sheets of
extended b(1S4) glycan chains and does not form a helical
H
O
NH
H2
C CH
NHCOCH3
H
C
O
H
H
H
H
H
H
O
O
O
H
OH
O
H
H
CH2OH
OH
H2C
H
O
H
OH
H
24. The unknown sugar is isomaltose.
CH2OH
O
H
OH H
H
OH
38. Each disaccharide unit of chondroitin sulfate has two negatively charged groups: a carboxylate group and a sulfate group.
One hundred of these disaccharide units would yield a net
charge of 2200.
40.
CH2OH
H
O
H
OH
H
H
O
HO
H
NH
C
H
CH
C
NH
C
R
O
O
CH3
R H (Ser)
R CH3 (Thr)
[From Schirm, M., Schoenhofen, I.C., Logan, S.M., Waldron,
K.C., and Thibault, P., Anal. Chem. 77, 7774–7782 (2005).]
Chapter 12 Solutions
2. The purple nonsulfur bacteria are photoheterotrophs.
They are similar to the photoautotrophs in that they can capture energy from sunlight but differ in that they are unable to
CHAPTER 12 Solutions | 27
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fix CO2. These bacteria are similar to the heterotrophs in that
an organic source of carbon is required. Thus the term photoheterotroph accurately describes the trophic strategy of this
organism.
4. Monosaccharides enter the cells lining the intestine via secondary active transport (see Fig. 9-15). Na1 ions and glucose
enter the cell in symport, both along their concentration gradients. The Na1 ions are subsequently pumped out of the cell
by the Na,K-ATPase transporter, which uses the free energy of
ATP hydrolysis to eject the Na1 ions against their concentration
gradient.
6. The pH optimum for pepsin is |2, which is the pH of
the stomach. The pH optimum for trypsin and chymotrypsin is |7–8, as the small intestine is slightly basic (see Table 2-3).
Each enzyme functions optimally in the conditions of its
environment.
8. The entry of glucose and amino acids into cells lining the
small intestine is accomplished by a secondary active transport process in which sodium ions are transported into the cell
along with the glucose or amino acid (see Fig. 9-15 and Problems 4 and 7). Adding an electrolyte such as sodium chloride
provides the sodium ions required for this cotransport. As the
intestinal cells import glucose and amino acids, water is also
absorbed.
10.
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will be protected from degradation by the destructive power
of these enzymes.
16. (a) oxidized
(b) reduced
(c) reduced
(d) oxidized
18. Plant-derived b-carotene is a precursor of vitamin A (see
Box 8-A) and is lipid-soluble. The b-carotene contained in the
salad and salsa ingredients dissolves in the fats contained in the
avocado, which enhances its absorption. [From Unlu, N.Z.,
Bohn, T., Clinton, S.K., and Schwartz, S.J., J. Nutr. 135, 431–
436 (2005)].
20. Individuals with gastrointestinal disorders might have a gastrointestinal tract that is not colonized by the appropriate vitamin B12–synthesizing bacteria. A deficiency in haptocorrin or
intrinsic factor would be manifested as a vitamin B12 deficiency,
since these proteins are essential for absorption of the vitamin.
Vegetarians and vegans who consume no animal products would
also be at risk for a deficiency of vitamin B12.
22. (a) Use Equation 12-1 to solve for the equilibrium constant, K eq:
K eq 5
[arginine ][ATP ]
[ phosphoarginine ][ADP ]
K eq 5
14.78 3 1023 2 13.87 3 1023 2
10.737 3 1023 2 10.750 3 1023 2
K eq 5 33.5
(b) Use Equation 12-2 to solve for DG 89:
O
H3C
(H2C)17
C
Cholesteryl ester
O
O
H3C
(H2C)17
C
Cholesteryl esterase
O
Fatty acid
HO
Cholesterol
DG °¿ 5 2RT ln K eq
DG °¿ 5 218.3145 3 1023 k J ? K21 ? mol21 2
1298 K2 ln 33.5
DG °¿ 5 28.7 kJ ? mol21
The reaction is spontaneous under standard conditions.
24. (a) See Sample Calculation 12-1. The equilibrium constant
can be derived by rearranging Equation 12-2:
K eq 5 e 2DG °¿/RT
21
23
21
21
K eq 5 e27.9 k J?mol /18.3145310 k J? K ? mol 2 1298 K2
K eq 5 e23.19
K eq 5 0.041
(b) At 378C, T 5 310 K:
12. A branched glycogen chain has many nonreducing ends
but only one reducing end. Enzymes specific for the nonreducing ends can act on many ends simultaneously, building up the
glycogen molecule when glucose is plentiful and degrading glycogen when glucose is in short supply. Enzymes that acted on
the reducing end would accomplish these processes much more
slowly.
14. If lysosomal hydrolytic enzymes function optimally at
pH 5, they will not function well at the cytosolic pH of 7.
Amino acid side chains essential for catalytic activity will become deprotonated at the higher pH. If lysosomal enzymes
leak out of the lysosome into the cytosol, cellular components
28
| CHAPTER 12 Solutions
DG 5 DG °¿ 1 RT ln
DG 5 7.9 kJ ? mol
21
DG 5 7.9 kJ ? mol21
[dihydroxyacetone phosphate]
[glyceraldehyde-3-phosphate]
1 18.3145 3 10 23 kJ ? K 21 ? mol 21 2
15 3 10 24 2
1310 K2 ln
11 3 10 24 2
21
1 4.1 kJ ? mol
DG 5 12 kJ ? mol21
(c) The reaction is not spontaneous as written. The reverse
reaction, in which DG 5 212 kJ ? mol21, would be spontaneous.
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26. The hydrolysis of pyrophosphate releases considerable free
energy (&33.5 kJ ? mol21). These two reactions are coupled, and
the overall DG value for the coupled reactions is negative; thus,
UDP–glucose formation occurs spontaneously.
28. (a)
ADP 1 Pi S ATP 1 H2O
DG °¿ 5 130.5 kJ ? mol 21
Under the given conditions, the reaction would produce
only 9.5 3 1028 M glucose-6-phosphate and thus is not a
feasible route to the production of this compound for the
glycolytic pathway.
(c)
4.184 kJ
1 mol
2000 kcal
3 0.33 5 90.5 moles ATP
3 1 day 3
3
1 kcal
day
30.5 kJ
(b) 90.5 moles ATP 3
505 g
mol
3
1 lb
5 20.8 lb
2200 g
(c) ATP does not accumulate but instead is constantly recycled. As ATP is used, its hydrolysis products, ADP and Pi ,
serve as reactants for ATP synthesis in the process of oxidative phosphorylation.
30. Reactions involving the phosphorylation of glucose (to glucose1-phosphate and glucose-6-phosphate) and glycerol (to glycerol-3phosphate) would require the hydrolysis of ATP to drive the
reaction, because transfer of a phosphoryl group from ATP
occurs with a greater change in free energy than the transfer of a
phosphoryl group to one of these compounds.
32. (a) Use Equation 12-2 to solve for DG 89:
DG °¿ 5 2RT ln K eq
DG °¿ 5 218.3145 3 1023 kJ ? K21 ? mol21 2 1298 K2 ln 0.41
DG °¿ 5 2.2 kJ ? mol21
K eq 5
Driving the reaction to the right using this method is not
feasible because it is impossible to achieve a concentration
of 13 M glucose inside the cell.
(d)
glucose 1 Pi z
y glucose-6-phosphate 1 H2O
DG °¿ 5 13.8 kJ ? mol21
ATP 1 H2O z
y ADP 1 Pi
DG °¿ 5 230.5 kJ ? mol21
glucose 1 ATP z
y glucose-6-phosphate 1 ADP
DG °¿ 5 216.7 kJ ? mol21
K eq 5 e2DG °¿/RT
21
K eq 5 e21 216.7 kJ ? mol
K eq 5 850
1310 K2 ln
15 3 1024 2
12.0 3 1023 2
DG 5 2.2 kJ ? mol21 2 3.57 kJ ? mol21
DG 5 21.37 kJ ? mol21
Under these conditions, the reaction will proceed as written.
34. (a) The equilibrium constant can be determined by rearranging Equation 12-2 (see Sample Calculation 12-1):
K eq 5 e2DG °¿/RT
213.8 k J ? mol21/ 18.314531023 k J ? K21 ? mol212 1298 K2
K eq 5 e
K eq 5 e25.57
K eq 5 0.0038
(b)
K eq 5
850 5
DG 5 2.2 kJ ? mol21 1 18.3145 3 1023 kJ ? K21 ? mol21 2
K eq 5
0.0038 5
0.0038 5
[glucose-6-phosphate]
[glucose ] [ Pi ]
[glucose-6-phosphate]
15.0 3 1023 2 15.0 3 1023 2
[glucose-6-phosphate]
12.5 3 1026 2
[glucose-6-phosphate ] 5 9.5 3 1028 M
2 / 18.314531023 kJ ? K21 ? mol212 1298 K2
K eq 5 e6.74
[fructose-6-phosphate ]
[glucose-6-phosphate ]
1250 3 1026 2
[glucose ] 15.0 3 1023 2
[glucose ] 5 13 M
(e)
DG 5 DG °¿ 1 RT ln
[ glucose ] [ Pi ]
0.0038 5
The reaction will proceed in the opposite direction as written.
(b)
[ glucose-6-phosphate]
[ glucose-6-phosphate ] [ADP ]
[glucose ] [ATP ]
1250 3 1026 2 11.25 3 1023 2
[glucose ] 15.0 3 1023 2
[ glucose ] 5 7.4 3 1028 M
(f ) The reaction can be accomplished at a much lower glucose concentration when the phosphorylation of glucose is
coupled to ATP hydrolysis (7.4 3 1028 M instead of 13 M).
This can be done because the second reaction couples the
phosphorylation of glucose with the exergonic hydrolysis of
ATP. Thus, an unfavorable reaction is converted to a favorable reaction.
36. I.
GAP z
y 1,3-BPG
1,3-BPG 1 H2O z
y 3PG 1 Pi
GAP 1 H2O z
y 3PG 1 Pi
DG °¿ 5 6.7 kJ ? mol21
DG °¿ 5 249.3 kJ ? mol21
DG °¿ 5 242.6 kJ ? mol21
II.
GAP z
y 1,3-BPG
1,3-BPG 1 ADP z
y 3PG 1 ATP
DG °¿ 5 6.7 kJ ? mol21
DG °¿ 5 218.8 kJ ? mol21
GAP 1 ADP z
y 3PG 1 ATP
DG °¿ 5 212.1 kJ ? mol21
The second scenario is more likely. The first coupled reaction
is more exergonic, but the second coupled reaction “captures”
some of this free energy in the form of ATP, which the cell
can use.
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38. (a) The value of C is determined from the slope of the
plot. The slope is equal to 1.9 3 1026 mM21 or 1.9 3
1023 M21.
4. (a)
2.2 kJ ? mol
1.2
[nick]/[phosphodiester] (10−4)
DG °¿ 5 2R T ln
21
5 218.3145 3 1023 kJ ? K21 ? mol21 2
[fructose-6-phosphate ]
1298 K2 ln
[ glucose-6-phosphate ]
20.88 5 ln
y (1.9 106) x 2.0 105
0.4
e20.88 5
0
10
20
30
40
0.41 5
50
[AMP] (mM)
(b)
[ glucose-6-phosphate ]
2.2 kJ ? mol21 5 22.47 kJ ? mol21 ln
0.8
0
[fructose-6-phosphate]
(b)
[ PPi ]
C5
K eq [ATP ]
K eq 5
[ glucose-6-phosphate ]
[fructose-6-phosphate ]
[ glucose-6-phosphate ]
[fructose-6-phosphate ]
[ glucose-6-phosphate ]
[fructose-6-phosphate ]
[ glucose-6-phosphate ]
DG 5 DG °¿ 1 RT ln
21.4 kJ ? mol
[PPi ]
K eq 5
C [ATP ]
21
5 2.2 kJ ? mol
11.9 3 1023 2 114 3 1026 2
4
(c) Use Equation 12-2 to solve for DG 89:
DG °¿ 5 2RT ln K eq
DG °¿ 5 218.3145 3 1023 kJ ? K21 ? mol21 2
1298 K2 ln 3.8 3 104
DG °¿ 5 226 kJ ? mol21
DG °¿ 5 ?
DG °¿ 5 248.5 kJ ? mol21
DG °¿ 5 226 kJ ? mol21
The DG 89 for the formation of the phosphodiester bond is
22.5 kJ ? mol21.
(e) The DG 89 value for the hydrolysis of a phosphodiester
bond in DNA is 222.5 kJ ? mol21, whereas the DG 89 value
for the hydrolysis of a typical phosphomonoester bond is
213.8 kJ ? mol21. Therefore, the phosphodiester bond in
DNA is less stable than the typical phosphomonoester
bond. [From Dickson, K., Burns, C. M., and Richardson,
J., J. Biol. Chem. 275, 15828–15831 (2000).]
Chapter 13 Solutions
2. Reactions 1, 3, and 10 are irreversible; the remaining reactions are reversible. Metabolically irreversible reactions are good
“control points” for a pathway.
| CHAPTER 13 Solutions
[ glucose-6-phosphate ]
21
23.6 kJ ? mol21 5 2.58 kJ ? mol21
K eq 5 3.8 3 10
30
[fructose-6-phosphate]
1 18.3145 3 10 23 kJ ? K21 ? mol21 2
[fructose-6-phosphate ]
1310 K2 ln
[ glucose-6-phosphate ]
11.0 3 1023 2
(d)
Nick z
y phosphodiester bond
z
ATP y AMP 1 PPi
ATP 1 nick z
y AMP 1 PPi
1 phosphodiester bond
[fructose-6-phosphate ]
ln
21.4 5 ln
e21.4 5
0.25 5
[fructose-6-phosphate ]
[ glucose-6-phosphate ]
[fructose-6-phosphate ]
[ glucose-6-phosphate ]
[fructose-6-phosphate ]
[ glucose-6-phosphate ]
[fructose-6-phosphate ]
[ glucose-6-phosphate ]
The reaction will proceed in the direction of fructose-6phosphate synthesis since DG , 0.
6. This would not be beneficial to the patient. In order to enter the glycolytic pathway, the glucose-6-phosphate would first
have to enter the cells. Glucose transporters recognize glucose,
not glucose-6-phosphate; thus, glucose-6-phosphate would be
unable to enter the cell for oxidation through the glycolytic
pathway.
8. (a) When F26BP binds to the allosteric site on PFK, one of
its phosphate groups may interact with the serine in some
manner, perhaps by hydrogen bonding. But if the serine is
replaced with the negatively charged aspartate, the negative
charges on the F26BP are repulsed and the F26BP cannot
bind to the enzyme.
(b) Apparently PFK cannot be fully active in the absence of
the F26BP allosteric activator, which plays an important role
in stimulating glycolysis. In its absence, the glycolytic pathway
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is less active, which accounts for the observed decrease in glucose consumption and ethanol production. [From Heinisch,
J. J., Boles, E., and Timpel, C., J. Biol. Chem. 271, 15928–
15933 (1996).]
10.
OH
C
H
O
C
CH2OPO2
3
Possible enediolate intermediate
12. (a) The cancer cells may express the GAPDH protein at
higher levels (i.e., transcription of the GAPDH gene and
translation of its mRNA may occur at a higher rate).
(b) The structure of GAPDH in cancer cells is probably different from the structure of GAPDH in normal cells. The
structure of the active site in GAPDH from cancer cells
might be altered in such a way that the binding of methylglyoxal is permitted, which then precludes the binding of
the substrate. Or the altered GAPDH might have a binding site for methylglyoxal elsewhere on the protein, which
causes a conformational change in the protein that alters
the substrate binding site so that the substrate can no longer
bind. [From Ray, M., Basu, N., and Ray, S., Mol. Cell.
Biochem. 177, 21–26 (1997).]
14. As the NADH/NAD1 ratio increases, the activity of
GAPDH decreases and less 1,3-bisphosphoglycerate is produced from glyceraldehyde-3-phosphate. NAD1 is a reactant
and NADH is a product of the reaction, so as NAD1 becomes
less available and NADH accumulates, the ratio of [product]/
[reactant] increases and the activity of the enzyme decreases.
16. Hexokinase-deficient erythrocytes have low levels of all glycolytic intermediates, since hexokinase catalyzes the first step of
glycolysis. Therefore, the concentration of 2,3-BPG in the erythrocyte will be decreased as well, favoring the oxygenated form of
hemoglobin and decreasing its p50 value. Pyruvate kinase–deficient
erythrocytes have high levels of 2,3-BPG since pyruvate kinase
catalyzes the last step of glycolysis. This blockade at the last step
causes the concentrations of all of the intermediates “ahead” of the
block to be increased. Thus the oxygen affinity of hemoglobin is
decreased with increased 2,3-BPG concentration, and the p50 value
increases as a result.
18. (a) In hepatocytes, the phospho-His on the phosphoglycerate mutase transfers its phosphate to the C2 position of
3PG to form 2,3-BPG. The [32P]-labeled phosphate on the
C3 position is transferred back to the enzyme to form the
2PG product, so initially the enzyme would be labeled. In
the next round of catalysis, the labeled phosphate on the enzyme is transferred to the C2 position of the next molecule
of 3PG substrate, so 2PG becomes labeled. Eventually, this
phosphate is transferred to ADP to form ATP, so ATP is
labeled.
(b) In the plant, the labeled phosphate is transferred to C2
to form 2PG, so 2PG is labeled and then eventually ATP.
The plant enzyme is not labeled.
20. (a) The glucose–lactate pathway releases 196 kJ ? mol21 of
free energy, enough theoretically to drive the synthesis of
196/30.5, or about 6, ATP.
(b) The complete oxidation of glucose releases 2850 kJ ?
mol21 of free energy, enough for 2850/30.5, or about 93,
ATP.
22. If hexokinase cannot be inhibited, all glucose that enters
the yeast cell will be phosphorylated at the expense of ATP to
produce glucose-6-phosphate. Glucose-6-phosphate is isomerized to fructose-6-phosphate, which is then phosphorylated to
fructose-1,6-bisphosphate, again at the expense of ATP. If concentrations of glucose are high, and if there is no mechanism to
inhibit hexokinase, then cellular ATP will be depleted in these
early steps. The early reactions of glycolysis proceed at a rate
greater than the rate of the later ATP-generating steps of glycolysis. ATP is therefore used faster than it is regenerated, and the
yeast mutants die as a result. [From Teusink, B., Walsh, M. C.,
van Dam, K., and Westerhoff, H. V., Trends Biochem. Sci. 23,
162–169 (1998).]
24. (a) When glucose is converted to pyruvate, NAD1 is reduced to NADH in the glyceraldehyde-3-phosphate dehydrogenase reaction. Without a mechanism for regenerating
NAD1, glycolysis could not continue.
(b) If the trypanosome reduced pyruvate to lactate, this step
would regenerate NAD1 and no other pathway for regenerating NAD1 would be needed.
(c) A drug that interfered with one of the enzymes in the
trypanosome’s NADH-oxidizing pathway would likely have
no effect on the host since mammals lack the enzymes that
would be inactivated by the drug.
100
26. Deamination of alanine produces pyruvate, a gluconeogenic
substrate, and deamination of aspartate produces oxaloacetate,
an intermediate of gluconeogenesis.
90
Hexokinase
deficient
Oxygen saturation (%)
80
70
60
Normal
erythrocytes
50
40
30
Pyruvate kinase
deficient
20
10
0
0
10
20
30
40
p O2 (torr)
50
60
28. Phosphoenolpyruvate carboxykinase catalyzes an essential step of gluconeogenesis. Lower expression of this enzyme decreases the gluconeogenic output of the liver, which
helps decrease the level of circulating glucose in patients with
diabetes.
30. Normally, muscle glycogen is degraded to glucose-6-phosphate, which enters glycolysis to be oxidized to yield ATP for
the active muscle. In anaerobic conditions, pyruvate, the end
CHAPTER 13 Solutions | 31
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product of glycolysis, is converted to lactate, which is released
from the muscle into the blood and enters the liver to be
converted back to glucose via gluconeogenesis. The patient’s
muscle cells are unable to degrade glycogen to glucose-6phosphate; thus, there is no glucose-6-phosphate to enter glycolysis and lactate formation does not occur. [From Stanbury,
J. B., Wyngaarden, J. B., and Fredrickson, D. S., The Metabolic Basis of Inherited Disease, pp. 151–153, McGraw-Hill,
New York (1978).]
32. In order for the enzyme to be active, the serine in the active
site must be phosphorylated. This phosphate group is donated
to C1 of the glucose-6-phosphate in the first step of the conversion of glucose-6-phosphate to glucose-1-phosphate. If glucose1,6-bisphosphate dissociates prematurely, the serine is not phosphorylated, the enzyme is not regenerated, and further rounds of
catalysis cannot occur.
34. Production of glucose-1-phosphate requires only an isomerization reaction catalyzed by phosphoglucomutase to convert it to glucose-6-phosphate, which can enter glycolysis. This
skips the hexokinase step and saves a molecule of ATP. Hydrolysis, which produces glucose, would require expenditure of an
ATP to phosphorylate glucose to glucose-6-phosphate.
36. (a) In a liver cell, glucose-6-phosphate has four possible
fates: It can be used to synthesize glycogen, it can be catabolized via glycolysis, it can be catabolized via the pentose
phosphate pathway, and it can be converted to glucose and
released from the cell.
(b) In a muscle cell, only the first three processes occur.
Muscle cells lack glucose-6-phosphatase and therefore
cannot release glucose from the cell.
38. (a) Since serum withdrawal decreases G6PD activity, serum
withdrawal should decrease the NADPH/NADP1 ratio.
NADPH is a product of the G6PD reaction and NADP1 is
a reactant, so decreased enzyme activity leads to an increase
in the oxidized form of the coenzyme and a decrease in the
reduced form.
(b) In the presence of DHEA, the NADPH/NADP1 ratio
should decrease, since DHEA inhibits G6PD activity, for
the reasons outlined in part (a).
(c) In the absence of any other factor, adding H2O2 should
not affect the ratio, since G6PDH can increase its activity
to produce more NADPH to react with the H2O2.
(d) In the absence of the serum containing growth factors, G6PD cannot handle the increased load of H2O2,
and the ratio would decrease. [From Tian, W.-N., Braunstein, L. D., Pang, J., Stuhlmeier, K. M., Xi, Q.-C., Tian,
X., and Stanton, R., J. Biol. Chem., 273, 10609–10617
(1998).]
40. If NADPH is used as a coenzyme for nitrate reduction,
it will become oxidized to NADP1 and will need to be regenerated. The pentose phosphate pathway produces NADPH,
so it is possible that the NADPH-producing enzymes of
this pathway, glucose-6-phosphate dehydrogenase and 632
| CHAPTER 14 Solutions
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phosphogluconate dehydrogenase, are stimulated by nitrate.
[From Hankinson, O., and Cove, D. J., J. Biol. Chem. 249,
2344–2353 (1974).]
42. G16BP inhibits hexokinase but stimulates PFK and pyruvate kinase. This means that glycolysis will be active, but only
if the substrate is glucose-6-phosphate, since glucose cannot be
phosphorylated in the absence of hexokinase activity. The pentose phosphate pathway is inactive, since 6-phosphogluconate
dehydrogenase is inhibited. Phosphoglucomutase is activated,
which converts glucose-1-phosphate (the product of glycogenolysis) to glucose-6-phosphate. Thus, in the presence of
G16BP, glycogenolysis is active and produces substrate for
glycolysis but not the pentose phosphate pathway. This is a
more efficient process than using glucose taken up from the
blood, which would need to be phosphorylated at the expense of ATP. [From Beitner, R., Trends Biol. Sci. 4, 228–230
(1979).]
44. (a) Inhibition of glycogen phosphorylase blocks glycogenolysis, and inhibition of the bisphosphatase blocks gluconeogenesis. As a result, the liver is unable to produce glucose
and the blood glucose level is low.
(b) The inhibition of fructose-1,6-bisphosphatase prevents gluconeogenesis even when substrates such as
glycerol and dihydroxyacetone phosphate are supplied.
Galactose relieves the hypoglycemia because it can be
phosphorylated, converted to glucose-6-phosphate, and
dephosphorylated to yield glucose that enters the circulation.
46. The concentration of xylulose-5-phosphate, an intermediate of the pentose phosphate pathway, increases following a
meal, when there is plenty of glucose being catabolized by the
pentose phosphate pathway. The increase in fructose-2,6-bisphosphate increases the flux through glycolysis while inhibiting flux through gluconeogenesis, so glycolysis produces large
amounts of pyruvate, which can be converted to acetyl-CoA.
At the same time, the production of lipid-synthesizing enzymes
increases, so the net result is that glucose is converted to acetylCoA and then to fat for storage.
Chapter 14 Solutions
2. The decarboxylation step is metabolically irreversible since
the CO2 product diffuses away from the enzyme. The other four
reactions are transfer reactions or oxidation–reduction reactions
(transfer of electrons) that are more easily reversed.
4. TPP is a cofactor in two of the enzymes associated with the
citric acid cycle—the pyruvate dehydrogenase complex and
the a-ketoglutarate dehydrogenase complex. If thiamine were
deficient, TPP would be deficient as well and the activities of
both these enzymes would decrease. As a result, the substrates
of these two reactions, pyruvate and a-ketoglutarate, would
accumulate.
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6.
O
R
O
C
O
CH3
N
C
C
S
CH3
Pyruvate
R
TPP
O
H
H
H
C
affinity for its substrate has decreased in the presence of the
inhibitor. The inhibitor competes with acetyl-CoA for binding to the citrate synthase active site. S-acetonyl-CoA can do
this because its structure resembles that of acetyl-CoA.
(c) Acetyl-CoA binds to pyruvate carboxylase and acts as
an activator. A thioester functional group must be required
for binding, since S-acetonyl-CoA, which lacks a thioester
functional group, cannot bind to this binding site. [From
Rubenstein, P., and Dryer, R., J. Biol. Chem. 255, 7858–
7862 (1980).]
1
4
CH3
Acetaldehyde
H R
O
H
O
CH3
N
C
C
CH3
R
S
CO2
R
HO
R
2
R
CH3
N
C
C
CH3
3
H
CH3
N
C
HO
S
O R
O
C
C
S
H 3C
C
S
H3C
R
CH3
N
HO
R
Resonance-stabilized carbanion
8. Citrate synthase acts on oxaloacetate (OAA) and fluoroacetylCoA, an acetyl-CoA analog, to produce fluorocitrate. The
fluorocitrate then serves as an inhibitor of aconitase. This
leads to an accumulation of citrate, since the citric acid cycle
can go no further if the aconitase reaction is inhibited.
O
FH2C
COO
C
FH2C
Fluoroacetate
S
CoA
Fluoroacetyl-CoA
citrate synthase
COO
CH
F
COO
C
HO
CH2
Oxaloacetate
COO
Fluorocitrate
10. (a)
O
Br
CH2
C
CH3
CoA
SH
O
HBr
H3C
C
CH2
S
CoA
S-Acetonyl-CoA
(b) S-acetonyl CoA is a competitive inhibitor. The Vmax of
the citrate synthase reaction is the same in the absence and
in the presence of the inhibitor. The K M has increased in
the presence of the inhibitor, indicating that the enzyme’s
12. Cis-aconitate is an intermediate in the reaction when citrate is converted to isocitrate by aconitase. Trans-aconitate
structurally resembles cis-aconitate and would be expected to
compete with cis-aconitate for binding to the enzyme. But because trans-aconitate is a noncompetitive inhibitor when citrate
is used as the substrate, the citrate binding site must be distinct
from the aconitate binding site. Citrate and trans-aconitate do
not compete for binding and can bind to the enzyme simultaneously, but when both substrate and inhibitor are bound, the
substrate cannot be converted to product. [From Villafranca, J.J.,
J. Biol. Chem. 249, 6149–6155 (1974).]
14. (a) When yeast use glucose as a carbon source, ATP is
obtained from the oxidation of glucose by glycolysis.
Glucose is oxidized anaerobically, without the citric acid
cycle. But when the food source is shifted from glucose
to acetate, yeast are required to obtain all of their energy
from the citric acid cycle and oxidative phosphorylation.
Acetate is converted to acetyl-CoA, which enters the citric acid cycle. In the absence of glycolysis, flux through
the citric acid cycle increases. Because isocitrate dehydrogenase is one of the main regulatory enzymes of the
cycle, increasing its expression increases the rate of its
reaction and increases the flux through the citric acid
cycle. In this manner, yeast can effectively use acetate as
a food source.
(b) For the wild-type yeast, the [NAD 1]/[NADH] ratio increases slightly as the cells switch from using glycolysis to using the citric acid cycle to generate ATP,
as described in part (a). There is an increase in flux
through the citric acid cycle, but perhaps the isocitrate
dehydrogenase cannot keep up with the demand, so
the ratio increases. There is a more dramatic change in
the ratio for the mutant because the isocitrate dehydrogenase enzyme is nonfunctional. This means that
the reaction, in which NAD 1 is a reactant and NADH
is a product, cannot occur, so the NAD1 reactant accumulates. These cells will eventually die if not given a
carbon source other than acetate. [From Minard, K. I.,
and McAlister-Henn, L., Arch. Biochem. Biophys. 483,
136–143 (2009).]
16. TPP attacks the carbonyl carbon of the succinyl phosphonate, but the subsequent step (the departure of the leaving
group) cannot occur because the COP bond cannot be broken.
A covalent bond forms between the enzyme and the inhibitor
(although the reaction is reversible, the reverse reaction occurs
slowly) and the substrate cannot bind.
CHAPTER 14 Solutions | 33
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H3C
/Users/user-f391/Desktop/15:03:10
H3C
N S
C
N S
O C
O
O
O
P
O
C
O
P
C
CH2
CH2
COO
O OH
CH2
CH2
COO
[From Bunik, V. I., Denton, T. T., Xu, H., Thompson, C. M.,
Cooper, A. J. L., and Gibson, G., Biochemistry 44, 10552–
10561 (2005).]
18. (a) Succinyl-CoA synthetase catalyzes the only substratelevel phosphorylation reaction in the citric acid cycle. The
enzyme with the ADP-specific b subunit could produce
ATP in the brain and muscle to meet the energy needs of
these tissues. The enzyme with the GDP-specific b subunit
could produce GTP needed by phosphoenolpyruvate carboxykinase for gluconeogenesis in the liver and kidneys.
(b) Individuals who lack a functioning a subunit cannot
carry out the succinyl-CoA synthetase reaction in any tissue, since this subunit is common to both forms of the
enzyme. Because aerobic respiration is impossible, the individual uses glycolysis followed by lactate fermentation to
obtain ATP. This doesn’t provide enough energy to sustain
life, so the person dies shortly after birth.
(c) When the gene for one of the b subunits is mutated, the
gene for the other subunit is normal. The individual would
suffer from decreased succinyl-CoA synthetase activity, but
operation of the citric acid cycle would allow the individual
to obtain some energy from aerobic respiration. However,
the individual’s ability to meet his/her energy needs is still
somewhat compromised, as shown by the elevated levels of
lactate and decreased life span.
20. Isocitrate lyase, a glyoxylate pathway enzyme, catalyzes the
conversion of isocitrate to succinate and glyoxylate. The glyoxylate product of this reaction goes on to form malate, which is
eventually converted to glucose via gluconeogenesis. The succinate product is not part of this pathway and is “disposed of ” by
mitochondrial succinate dehydrogenase, which regenerates the
oxaloacetate needed to keep the glyoxylate pathway operating.
22. Levels of pyruvate, lactate, and fumarate would all increase
in the patient. The concentration of malate would decrease. Pyruvate and lactate increase because the patient relies on glycolysis and lactate fermentation in the absence of proper citric acid
cycle function. The concentration of fumarate increases because
it cannot be converted to malate in the absence of fumarase. This
also explains why the concentration of malate decreases.
24. When cells obtain energy in the absence of oxygen, glucose is oxidized to pyruvate, which is subsequently reduced to
lactate with concomitant regeneration of NAD1. The citric acid
cycle is not active, and the cells respond by down-regulating the
activity of the citric acid cycle enzyme malate dehydrogenase.
34
| CHAPTER 14 Solutions
In the presence of oxygen, pyruvate is converted to acetyl-CoA,
which enters the citric acid cycle so that ATP can be produced
by oxidative phosphorylation. Under aerobic conditions, malate
dehydrogenase is absolutely essential for the regeneration of oxaloacetate as part of the citric acid cycle; therefore, activity levels
are much higher.
26. The volume increase that occurs when the bread dough rises
is due to the CO2 produced by the anaerobic oxidation of glucose. Sugar (sucrose) in the bread dough is hydrolyzed to glucose
and fructose by enzymes in the yeast. Then the fructose and glucose both enter glycolysis. The end product is pyruvate, which
is converted to acetaldehyde. In this step, a CO2 is released. The
acetaldehyde is then reduced to ethanol, which evaporates when
the bread is baked.
28. The alternate pathway bypasses the succinyl-CoA synthetase reaction of the standard citric acid cycle, a step that is accompanied by the phosphorylation of a nucleoside diphosphate.
The alternative pathway therefore generates one less nucleoside
triphosphate than the standard citric acid cycle. There is no difference in the number of reduced cofactors generated.
30. In the reversible reaction shown, the amino acid aspartate
and the glycolytic product pyruvate can undergo a transamination in which aspartate’s amino group is transferred, leaving oxaloacetate, which is a citric acid cycle intermediate.
32. An increase in the concentration of glucose increases flux
through glycolysis, which produces more pyruvate. The pyruvate dehydrogenase complex activity increases in response to
the increase in pyruvate so that the conversion of pyruvate to
acetyl-CoA can increase proportionally. At the same time, more
oxaloacetate is required, since equimolar amounts of acetyl-CoA
and oxaloacetate are required for the first step of the citric acid
cycle. Therefore, the activity of pyruvate carboxylase, the enzyme
that catalyzes the conversion of pyruvate to oxaloacetate, also
increases.
34. Glutamine can be converted to glutamate, which can be
converted to a-ketoglutarate. This citric acid cycle intermediate is eventually converted to oxaloacetate, so the glutamine is
a source of the oxaloacetate needed for gluconeogenesis when
pyruvate carboxylase is not functioning. The oxaloacetate derived from glutamine can also replenish citric acid cycle intermediates that are normally produced by the pyruvate carboxylase reaction.
36. The exercising muscle required greater concentrations of
ATP to power it, so the rates of glycolysis and the citric acid
cycle increased. Phosphoenolpyruvate was converted to pyruvate
more rapidly, so the concentration of phosphoenolpyruvate decreased. Some of the pyruvate was converted to acetyl-CoA via
the pyruvate dehydrogenase reaction. Since equimolar amounts
of acetyl-CoA and oxaloacetate are required for the first step
of the citric acid cycle, some of the pyruvate was converted to
oxaloacetate via the pyruvate carboxylase reaction. This explains
why oxaloacetate concentrations increased. The concentration of
pyruvate did not increase because a steady state was reached: The
rate of production of pyruvate from phosphoenolpyruvate was
equal to the rate of consumption of pyruvate.
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38. Ethanol is converted to acetaldehyde and then to acetate, and
acetate is converted to acetyl-CoA. Acetyl-CoA then enters the
glyoxylate pathway. The first step is the synthesis of citrate from
acetyl-CoA and oxaloacetate. Citrate is isomerized to isocitrate.
Isocitrate lyase then splits isocitrate into succinate and glyoxylate.
Succinate leaves the glyoxysome and enters the mitochondrion,
where it participates in the citric acid cycle. The glyoxylate fuses
with a second molecule of acetyl-CoA to yield malate. Malate
leaves the glyoxysome and enters the cytosol, where it is converted
to oxaloacetate via the malate dehydrogenase reaction. Oxaloacetate can then enter gluconeogenesis to form glucose.
40. The mutant cannot convert isocitrate to a-ketoglutarate, so
the citric acid cycle is blocked at this point. Adding glutamate
to the culture medium bypasses the block, because glutamate
can be converted to a-ketoglutarate and the citric acid cycle can
continue.
42. (a) Hydroxycitrate differs by the addition of a hydroxyl group.
(b) Based on its structural similarity to the substrate citrate, hydroxycitrate is most likely to act as a competitive
inhibitor.
(c) Acetyl-CoA synthesized in the mitochondria from pyruvate, the product of carbohydrate catabolism, is made
available in the cytosol by the action of ATP-citrate lyase
(see Fig. 14-16). Inhibiting this enzyme might decrease the
amount of acetyl-CoA available for fatty acid synthesis.
(d) Since cholesterol synthesis also begins with acetyl-CoA,
hydroxycitrate might reduce the production of cholesterol
and the structurally related steroid hormones. (Although
hydroxycitrate inhibits ATP-citrate lyase in vitro, controlled
clinical trials have shown that the compound has no significant effect on weight loss in humans.)
44. (a) The acetyl-CoA enters the glyoxylate pathway to produce malate, which can then be converted to oxaloacetate
and used to synthesize glucose. Acetyl-CoA can also condense with oxaloacetate to form citrate, which can be converted to isocitrate and then to a-ketoglutarate. Glutamate
dehydrogenase converts a-ketoglutarate to glutamate by
reductive amination. Alternatively, a-ketoglutarate can be
converted to glutamate by transamination.
(b) Aspartate undergoes transmination to form oxaloacetate, which can condense with acetyl-CoA to form citrate.
Citrate is then converted to isocitrate, which is converted to
a-ketoglutarate. a-Ketoglutarate can then be converted to
glutamate.
46. (a) Isocitrate lyase and malate synthase are found in the glyoxysome, so the pathogen is using the glyoxylate pathway.
The glyoxylate pathway is linked to the citric acid cycle,
so the citrate synthase and malate dehydrogenase enzymes
are upregulated as well. The glyoxylate pathway allows the
pathogen to use two-carbon sources to synthesize malate,
and from there, glucose and larger biological macromolecules. The phagosome is not likely to be a rich source of glucose and amino acids that the pathogen can use for biosynthesis, so the glyoxylate pathway is essential to its survival.
(b) The host does not have the enzymes found in the glyoxylate pathway, so inhibitors of isocitrate lyase and malate
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synthase would be good targets, assuming that there is a way
to deliver these inhibitors to the phagosome. [From Lorenz,
M. C., and Fink, G. R., Euk. Cell 1, 657–662 (2002).]
48. (a)
PEP
CO2
PEP carboxylase
OAA
NADH
NAD
malate dehydrogenase
malate
H2O
fumarase
fumarate
QH2
succinate dehydrogenase
Q
succinate
(b) If the bacteria were allowed to grow aerobically, the carbon atoms of fuel molecules would be completely oxidized
to carbon dioxide. The citric acid cycle would function in a
forward, catabolic direction. Depriving the bacteria of oxygen ensures that the final three steps of the citric acid cycle
will function in a reverse, anabolic direction to produce
the desired product, succinate. [From Agarwal, L., Isar, J.,
Meghwanshi, G. K., and Saxena, R. K., Enzyme and Microbial Technology 40, 629–636 (2007).]
Chapter 15 Solutions
2. Reverse the cytochrome a3 half-reaction and the sign of its
E89 value to indicate oxidation, multiply the coefficients by 2
so that the number of electrons transferred will be equal, then
combine the half-reactions and their E89 values.
2 cyt a3 (Fe21) S 2 cyt a3 (Fe31) 1 2 e2
1
2
½ O2 1 2 H 1 2 e
21
E89 5 20.385 V
S H2O
E89 5 0.815 V
1
DE89 5 0.430 V
2 cyt a3 (Fe ) 1 ½ O2 1 2 H S
2 cyt a3 (Fe31) 1 H2O
Use Equation 15-4 to calculate DG 89 for this reaction:
DG °¿ 5 2nFDE°¿
DG °¿ 5 2122 196,485 J ? V21 ? mol21 2 10.430 V2
DG °¿ 5 283 kJ ? mol21
The reaction is spontaneous under standard conditions.
CHAPTER 15 Solutions | 35
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4. Consult Table 15-1 for the relevant half-reactions involving
acetoacetate and NADH. Reverse the NADH half-reaction and
the sign of its E89 value to indicate oxidation, then combine the
half-reactions and their E89 values.
acetoacetate 1 2 H1 1 2 e2 S 3-hydroxybutyrate
NADH S NAD1 1 H1 1 2 e2
E89 5 20.346 V
E89 5 0.315 V
acetoacetate 1 NADH 1 H1 S NAD1 1 3-hydroxybutyrate DE89 5 20.031 V
Use Equation 15-4 to calculate DG 89 for this reaction:
DG °¿ 5 2nFDE°¿
DG °¿ 5 2122 196,485 J ? V21 ? mol21 2 120.031 V2
DG °¿ 5 6.0 kJ ? mol 21
Consult Table 15-1 for the relevant half-reactions involving
acetoacetate and FADH2. Reverse the FADH2 half-reaction and
the sign of its E89 value to indicate oxidation, then combine the
half-reactions and their E89 values.
1
2
acetoacetate 1 2 H 1 2 e S 3-hydroxybutyrate
E89 5 20.346 V
E89 5 0.0 V
FADH2 S FAD 1 2 H1 1 2 e2
acetoacetate 1 FADH2 S FAD 1 3-hydroxybutyrate DE89 5 20.346 V
Use Equation 15-4 to calculate DG 89 for this reaction:
DG °¿ 5 2nFDE°¿
DG °¿ 5 2122 196,485 J ? V21 ? mol21 2 120.346 V2
DG °¿ 5 66.8 kJ ? mol21
The reduction of acetoacetate by NADH is more favorable than
reduction by FADH2, as shown by the DG89 values calculated
above. Although neither reaction is spontaneous under standard conditions, reduction by FADH2 is far more unfavorable.
The reduction of acetoacetate by NADH is, in fact, a favorable
process under cellular conditions, which differ from standard
conditions.
6. Consult Table 15-1 for the relevant half-reactions involving
succinate and NAD1. Reverse the succinate half-reaction and
the sign of its E89 value to indicate oxidation, then combine the
half-reactions and their E89 values.
succinate S fumarate 1 2 H1 1 2 e2
E89 5 20.031 V
NAD1 1 H1 1 2 e2 S NADH
E89 5 20.315 V
succinate 1 NAD1 1 H1 S NADH 1 fumarate DE89 5 20.346 V
Use Equation 15-4 to calculate DG 89 for this reaction:
DG °¿ 5 2nFDE°¿
DG °¿ 5 2122 196,485 J ? V21 ? mol21 2 120.346 V2
DG °¿ 5 66.8 kJ ? mol21
Consult Table 15-1 for the relevant half-reactions involving succinate and FAD. Reverse the succinate half-reaction and the sign
of its E89 value to indicate oxidation, then combine the halfreactions and their E89 values.
succinate S fumarate 1 2 H1 1 2 e2
E89 5 20.031 V
FAD 1 2 H1 1 2 e2 S FADH2
E89 5 0.0 V
succinate 1 FAD S fumarate 1 FADH2 DE89 5 20.031 V
Use Equation 15-4 to calculate DG 89 for this reaction:
DG °¿ 5 2n FDE°¿
DG °¿ 5 2122 196,485 J ? V21 ? mol21 2 120.031 V2
DG °¿ 5 6.0 kJ ? mol21
36
| CHAPTER 15 Solutions
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The oxidation of succinate by FAD is more favorable than oxidation by NAD1, as shown by the DG 89 values calculated above.
Although neither reaction is spontaneous under standard conditions, oxidation by NAD1 is far more unfavorable. The oxidation of succinate by FAD is, in fact, a favorable process under
cellular conditions (citric acid cycle, Section 14-2), which differ
from standard conditions.
8. Reverse the half-reaction for the iron–sulfur protein to indicate that it is being oxidized. Add the two half-reactions to
obtain the DE89 for the reaction.
FeS(red ) S FeS(ox) 1 e2
cyt c1 (Fe31) 1 e2 S cyt c1 (Fe21)
FeS (red ) 1 cyt c1 (Fe31) S FeS(ox) 1 cyt c1 (Fe21)
E89 5 20.280 V
E89 5 0.215 V
DE89 5 20.065 V
Use Equation 15-4 to calculate DG 89 for this reaction:
DG °¿ 5 2nFDE°¿
DG °¿ 5 2112 196,485 J ? V21 ? mol21 2 120.065 V2
DG °¿ 5 6.27 kJ ? mol21
The positive DG 89 indicates that the electron transfer is unfavorable under standard conditions. However, cellular conditions are
not necessarily standard conditions, and the DG for this reaction
is likely to be negative. Also, since this reaction occurs as part of
the electron transport chain, the electrons gained by cytochrome
c1 will be passed along to Complex IV, in effect coupling the two
reactions, which would also tend to make the process more favorable than the DG 89 indicates.
10. (a) Consult Table 15-1 for the relevant half-reactions involving NADH and coenzyme Q. Reverse the NADH halfreaction and the sign of its E89 value to indicate oxidation,
then combine the half-reactions and their E89 values.
Q 1 2 H1 1 2 e2 S QH2
E89 5 0.045 V
NADH S NAD1 1 H1 1 2 e2
E89 5 0.315 V
NADH 1 Q 1 H1 S NAD1 1 QH2
DE895 0.360 V
Use Equation 15-4 to calculate DG 89 for this reaction:
DG °¿ 5 2nFDE°¿
DG °¿ 5 2122 196,485 J ? V21 ? mol21 2 10.360 V2
DG °¿ 5 269.5 kJ ? mol21
The phosphorylation of ADP to ATP requires 130.5 kJ ?
mol21 of free energy. Assuming that this reaction is 35% efficient (see Problem 9), 0.8 mol ATP can be synthesized under
standard conditions.
69.5 kJ ? mol21
10.352 5 0.80 mol
30.5 kJ ? mol21
ÿ
(b) Consult Table 15-1 for the relevant half-reactions involving ubiquinol and cytochrome c. Reverse the ubiquinol halfreaction and the sign of its E89 value to indicate oxidation,
multiply the coefficients in the cytochrome c equation by
2 so that the number of electrons transferred will be equal,
then combine the half-reactions and their E89 values.
QH2 S Q 1 2 H1 1 2 e2
2 cyt c (Fe31) 1 2 e2 S 2 cyt c (Fe21)
QH2 1 2 cyt c (Fe31) S Q 1 2 H1 1 2 cyt c (Fe21)
E89 5 20.045 V
E89 5 0.235 V
DE89 5 0.190 V
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Use Equation 15-4 to calculate DG 89 for this reaction:
DG °¿ 5 2nFDE°¿
DG °¿ 5 2122 196,485 J ? V21 ? mol21 2 10.190 V2
DG °¿ 5 236.7 kJ ? mol21
The phosphorylation of ADP to ATP requires 130.5 kJ/mol
of free energy. Assuming that this reaction is 35% efficient,
0.42 mol ATP can be synthesized.
36.7 kJ ? mol21
10.352 5 0.42 mol
30.5 kJ ? mol21
ÿ
(c) Consult Table 15-1 for the relevant half-reactions
involving cytochrome c and oxygen. Reverse the cytochrome
c half-reaction and the sign of its E89 value to indicate oxidation, multiply the coefficients by 2 so that the number of
electrons transferred will be equal, then combine the halfreactions and their E89 values.
2 cyt c (Fe21) 1 2 e2 S 2 cyt c (Fe31)
E89 5 20.235 V
½ O2 1 2 H1 1 2 e2 S H2O
21
1
E89 5 0.815 V
31
2 cyt c (Fe ) 1 ½ O2 1 2 H S 2 cyt c (Fe ) 1 H2O
DE89 5 0.580 V
Use Equation 15-4 to calculate DG 89 for this reaction:
DG °¿ 5 2nFDE°¿
DG °¿ 5 2122 196,485 J ? V 21 ? mol 21 2 10.580 V2
DG °¿ 5 2112 kJ ? mol 21
The phosphorylation of ADP to ATP requires 130.5 kJ/
mol of free energy. Assuming that this reaction is 35% efficient, 1.3 mol ATP can be synthesized.
112 kJ ? mol 21
10.352 5 1.3 mol
30.5 kJ ? mol 21
12. This is not enough free energy to drive ATP synthesis under
standard conditions (DG 89 5 30.5 kJ ? mol21).
14. Adding succinate to rotenone-blocked mitochondria
effectively bypasses the block as succinate donates its electrons
to ubiquinone and electron transport resumes. Adding succinate
is not an effective bypass for antimycin A2 or cyanide-blocked
mitochondria because succinate donates its electrons upstream
of the block.
16. The donation of a pair of electrons to cytochrome c and
then to Complex IV will result in the synthesis of about 1.3
ATP per atom of oxygen (½ O2). Therefore, the P:O ratio of this
compound is 1.3 [see Solution 10(c)].
18. Myxothiazol inhibits electron transfer in Complex III (specifically, it inhibits the transfer of electrons from reduced ubiquinol to both cytochrome b and the iron–sulfur protein redox
centers in Complex III).
20. (a) Pyruvate and malate are oxidized by pyruvate dehydrogenase and malate dehydrogenase, respectively, and
both reactions produce NADH, which can enter electron
transport.
(b) Fluoxetine inhibits electron transport generally, since
the rate of electron transport falls from an average of 163 to
77 when 0.15 mM fluoxetine is present. The inhibition is
primarily at Complex I, since the rate of electron transport
in the presence of fluoxetine is not decreased substantially
when succinate (which donates its electrons to Complex II)
and rotenone (which inhibits electron transfer in Complex I)
are added. However, fluoxetine also has the ability to inhibit Complex IV somewhat, since electron transport in the
presence of fluoxetine decreases in the presence of ascorbate
(which donates its electrons to cytochrome c) and TMPD
(which donates its electrons to Complex IV).
(c) By decreasing both the rate of electron transport and ATP
synthesis, fluoxetine decreases the rate of ATP production in
the brain. The brain relies on a constant source of ATP for
proper function, so decreased ATP production could lead
to an impairment of brain function. [From Curti, C., Mingatto, F. E., Polizello, A. C. M., Galastri, L. O., Uyemura,
S. A., and Santos, A. C., Mol. Cell. Biochem. 199, 103–109
(1999).]
22. Cytochrome c is a water-soluble, peripheral membrane
protein and is easily dissociated from the membrane by adding salt solutions that interfere with the ionic interactions that
tether it to the inner mitochondrial membrane. Cytochrome c1
is an integral membrane protein and is largely water-insoluble
due to the nonpolar amino acids that interact with the acyl
chains of the membrane lipids. Detergents are required to dissociate cytochrome c1 from the membrane because amphiphilic
detergents can disrupt the membrane and coat membrane proteins, acting as substitute lipids in the solubilization process.
24. The molasses and oil are food for microorganisms. As the
food is consumed, the rates of respiration and oxygen consumption increase. Eventually, the depletion of oxygen creates a more
reducing environment that favors the reduction of Cr(VI) compounds to Cr(III) compounds.
26. Use the rearrangement of Equation 15-7 as shown in Sample Calculation 15-2.
DG 5 2.303RT 1pHin 2 pHout 2 1 ZFDc
DG 5 2.30318.3145 3 10 23 J ? K 21 ? mol 21 2 1310 K2
17.6 2 7.22 1 112 196,485 J ? V 21 ? mol 21 2 10.200 V2
DG 5 2.4 kJ ? mol 21 1 19.3 kJ ? mol 21
DG 5 21.7 kJ ? mol 21
ÿ
28. (a) The impermeability of the membrane to ions is an important feature of chemiosmosis, because the free movement of ions other than H1 would dissipate the electrical
component of the proton gradient and thereby compromise
the ability of the gradient to supply free energy for ATP
synthesis.
(b) If the membrane were permeable to other ions, the
proton gradient would be a simple chemical gradient
rather than an electrochemical gradient. However, it
would still be able to serve as a source of free energy for
ATP synthesis.
30. A phosphate ion enters the mitochondrial matrix in symport with a proton. This proton moves in a direction opposite
to the protons translocated by Complexes I, II, and IV. Thus,
the difference in pH between the mitochondrial matrix and the
intermembrane space becomes smaller.
CHAPTER 15 Solutions | 37
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32. (a) Aerobic respiration yields 32 ATP (see Problem 31),
each of which requires 30.5 kJ/mol to synthesize:
32 ATP 3 30.5 kJ ? mol21ATP
3 100 5 34%
2850 kJ ? mol21
ÿ
Lactate fermentation yields 2 ATP:
2 ATP 3 30.5 kJ ? mol21ATP
3 100 5 31%
196 kJ ? mol21
ÿ
Alcoholic fermentation yields 2 ATP:
2 ATP 3 30.5 kJ ? mol21ATP
3 100 5 26%
235 kJ ? mol21
(b) Organisms that can oxidize glucose in the presence of
oxygen have an advantage over anaerobic organisms because
these organisms can extract more energy per glucose. This
may have been important in evolution.
34. Stimulation of the glycolytic enzymes hexokinase and phosphofructokinase increases glycolytic flux and allows the yeast
cells to obtain ATP more quickly under anaerobic conditions.
[From Beitner, R., Cohen, T. J., Nordenberg, J., and Haberman,
S., Biochim. Biophys. Acta 586, 266–277 (1979).]
36. If the translocase is unable to function, ATP will not
be able to exit the mitochondrial matrix and ADP will not
be transported inside. Without ADP, the substrate for ATP
synthase, ATP synthesis will not occur. Since electron transport and oxidative phosphorylation are coupled, a decrease in
the rate of oxidative phosphorylation will decrease the rate of
electron transport.
38. (a) ATP synthesis decreases dramatically in the presence of
oligomycin, since proton transfer is required to stimulate
rotation of the g subunit of ATP synthase, which causes
the sequential conformational change of the b subunits that
catalyze the phosphorylation of ADP to ATP.
(b) Since oxidative phosphorylation and electron transport
are coupled, a decrease in the rate of oxidative phosphorylation will also affect the rate of electron transport. If ATP
synthesis is not occurring, the proton gradient is not “discharged” and the rate of electron transport decreases.
(c) A decrease in the rate of electron transport will also decrease the rate of oxygen consumption.
(d) If dinitrophenol is added, the proton gradient is dissipated, or “discharged,” but not in a way that leads to ATP
synthesis. Therefore, ATP synthesis still does not occur, but
electron transport and oxygen consumption resume, with
the free energy of the process released as heat.
40. FCCP is a lipid-soluble compound that can pass through
both the outer and inner mitochondrial membranes and into
the matrix. FCCP has an acidic hydrogen on the central nitrogen that dissociates in the basic environment of the matrix. The FCCP passes back through the inner membrane to
the intermembrane space, where it is reprotonated and the
entire cycle begins again. In this manner, the FCCP shuttles
protons into the mitochondrial matrix and dissipates the proton gradient.
38
| CHAPTER 15 Solutions
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42. (a) The P:O ratio is decreased. For some reason, electron
transport and oxidative phosphorylation have been uncoupled. NADH is oxidized in the electron transport chain,
oxygen is reduced to water, but ATP may not be synthesized
if ADP is not available. This decreases the P:O ratio because
oxygen consumption occurs to a greater extent than ADP
phosphorylation.
(b) If the energy released in electron transport is not used
to synthesize ATP, it is released as heat, which accounts
for the elevated body temperature. If sufficient ATP is not
synthesized to meet energy needs, the rate of electron transport increases, which leads to an increased consumption of
O2 and an increase in the concentration of reduced coenzymes that enter electron transport, thus increasing metabolic rate.
(c) The patient will not be able to carry out strenuous
exercise under aerobic conditions because her muscle cells
are incapable of generating enough ATP to power the
muscle.
44. Since it has 9 c subunits, the bacterial enzyme can theoretically produce 3 ATP for every 9 protons translocated, or one
ATP per 3 H1. In the chloroplast, 3 ATP are synthesized for
every 14 protons, or 1 ATP per 4.7 H1. Thus, the bacterium is
more efficient in its use of the proton gradient established during
electron transport and has a higher ratio of ATP produced per
oxygen consumed.
46. During anaerobic fermentation, only 2 ATP molecules can
be obtained from each glucose molecule. Thus, in order to get
enough ATP to satisfy the energy needs of the cell, glucose consumption by the glycolytic pathway predominates, and only a
small amount is left over for the pentose phosphate pathway.
But during aerobic respiration, 32 ATP are available per glucose
molecule. Thus, less glucose is needed to synthesize the same
amount of ATP, so a larger fraction of glucose can enter the pentose phosphate pathway.
48. (a) In the root system of the skunk cabbage, starch is broken
down to yield glucose, which is catabolized via glycolysis, the
citric acid cycle, and the electron transport chain. The oxidation of glucose provides the reduced cofactors (NADH
and QH2) required to keep electron transport going so that
thermogenesis can occur.
(b) In the skunk cabbage, thermogenesis increases as the
temperature decreases. Thus, the rate of aerobic oxidation
of glucose increases to increase the flux of NADH and QH2
through the electron transport chain. Since oxygen is the
final electron acceptor, an increase in electron transport will
also increase oxygen consumption. When the ambient air
temperature is warmer, the need for heat production is less,
so there is less flux through the electron transport chain.
Thus, oxygen consumption decreases during the day.
(c) The synthesis of the uncoupling protein increases with
decreasing temperature, presumably by an increase in transcription of the mRNA that codes for the uncoupling protein (although a decrease in the rate of mRNA degradation
would produce the same results). The increased amount of
mRNA likely results in an increase in concentration of the
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uncoupling protein so that the mitochondrial proton gradient
would be discharged. Thus, at low temperatures, the potato
could generate heat rather than ATP. [From Laloi, M., Klein,
M., Riesmeier, J.W., Müller-Röber, B., Fleury, C., Bouillaud,
F., and Ricquier, D., Nature 389, 135–136 (1997); Knutson,
R. M., Science 186, 746–747 (1974); and Seymour, R. S.,
and Schultze-Motel, P., Nature 383, 305 (1996).]
50. (a) Substrate-level phosphorylation is catalyzed by phosphoglycerate kinase and pyruvate kinase in glycolysis, and
by succinyl-CoA synthetase in the citric acid cycle.
(b) C6H12O6 1 6 O2 S 6 CO2 1 6 H2O
Chapter 16 Solutions
2. Both cellular organelles are membrane-bound structures
with their own DNA that codes for proteins essential for organelle function (respiration for the mitochondrion, photosynthesis for the chloroplast). Both organelles have an inner and
an outer membrane that enclose an intermembrane space. The
outer membrane is fairly porous in both organelles, whereas the
inner membrane is fairly impermeable. The inner compartment
of the mitochondrion is referred to as the mitochondrial matrix,
while the inner compartment of the chloroplast is called the
stroma. Contained within the stroma is a third membranous
structure called the thylakoid, which is analogous to mitochondrial inner-membrane cristae. The thylakoid membrane and the
cristae differ in shape; the thylakoid consists of stacks of flattened vesicles, whereas the cristae are tubular or highly folded.
4. Lipids with highly unsaturated fatty acids have low melting
points, since the cis double bonds create bends that make it difficult for the lipids to pack together. Because there are few intermolecular forces binding these lipids together, they generate
a highly fluid membrane, which must be essential for photosynthesis in some way.
6. (a) If the synthesis of each ATP requires 30.5 kJ ? mol21,
then 9.8 mol ATP (300/30.5) could be synthesized.
(b) About 5.6 mol ATP (170/30.5) could be synthesized.
8. Photosynthetic bacteria can absorb light in the infrared and
ultraviolet regions of the electromagnetic spectrum.
10. (a) The sequences would provide only limited information, since even highly homologous proteins such as myoglobin and hemoglobin (which have very similar structures) have limited sequence identity (see Section 5-1).
The three-dimensional structures of the proteins would be
more likely to indicate whether they function similarly.
(b) Low-resolution models that included the heme groups
would be more useful, since the positions of the heme
groups within the proteins might indicate whether they
function similarly in accepting and donating electrons to
their redox partners. It is possible that the heme groups
could have similar orientations even if the apoproteins did
not resemble each other in overall tertiary structure.
12. Photooxidation would not be a good protective mechanism
since it might interfere with the normal redox balance among the
electron-carrying groups in the thylakoid membrane. Releasing the
energy by exciton transfer or fluorescence (emitting light of a longer
/Users/user-f391/Desktop/16:03:10
wavelength) could potentially funnel light energy back to the overactive photosystems. Dissipation of excess energy as heat would be
the safest mechanism, since the photosystems do not have any way
to harvest thermal energy to drive chemical reactions.
14. If the photosystems were in close proximity, then Photosystem
II might not undergo photooxidation and instead act as a lightharvesting complex for Photosystem I. Because the reaction center
of Photosystem II absorbs light with a peak wavelength of 680 nm,
it could pass energy to the reaction center of Photosystem I, which
absorbs lower-energy (longer-wavelength) light at its P700 group.
16. Myxothiazol inhibits electron transfer in the cytochrome
b6 f complex.
18. The alanine residue at position 251 is essential for photosynthesis and photoautotrophic growth and may be part of
the binding site for herbicides. When the alanine is mutated
to a cysteine, which resembles alanine except that a sulfhydryl group has replaced a hydrogen, the mutated protein is
similar to the wild-type. When the alanine is changed to an
amino acids that is similar in size but more polar (such as Gly,
Pro, or Ser), photosynthesis is impaired. Replacement with a
larger, nonpolar amino acid results in a mutant that is further
impaired, and replacement with a larger polar amino acid results in an organism that is not photosynthetically competent.
Therefore, the residue at position 251 must be relatively small
and nonpolar. Residues of increasing size and polarity result
in photosynthetically impaired organisms. [From Lardans, A.,
Förster, B., Prásil, O., Falkowski, P. G., Sobolev, V., Edelman,
M., Osmond, C. B., Gillman, N. W., and Boynton, J. E., J. Biol.
Chem. 273, 11082–11091 (1998).]
20. DG 5 2.303RT 1pHin 2 pHout 2 1 ZFDc
DG 5 2.303 18.3145 J ? K21 ? mol21 2 1298 K2 10.752
1 112 196,485 J ? V21 ? mol21 2 10.20 V2
DG 5 4300 J ? mol21 1 19,300 J ? mol21
DG 5 23.6 kJ ? mol21
For both organelles, the translocation is endergonic. In the chloroplast, most of this energy is due to the concentration (pH) difference on the two sides of the membrane; in the mitochondrion,
most of the energy is due to the membrane potential (Dc).
22. Use Equation 16-1 and multiply by 2 to calculate the energy
of the photons:
E5
E5
hc
32
l
16.626 3 10234 J ? s2 12.998 3 108 m ? s21 2 122
6 3 1027 m
219
E 5 6.6 3 10
J
The energy of two photons is nearly twofold greater than the
energy change for the oxidation of one molecule of water by
NADP1, so, in theory, the two photons supply sufficient energy
to drive the oxidation.
24. Some of the label appears in molecular oxygen released by
the water-splitting reaction of Photosystem II. Some of the label
appears in other compounds, including intermediates of the Calvin
cycle, since H2O participates in the rubisco reaction (Fig. 16-24).
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26. Complex III of the mitochondrial electron transport chain
is analogous to the cytochrome b6 f complex. In the presence
of antimycin A, electron flow in the cytochrome b6 f complex
would be inhibited. Electrons would not reach Photosystem I,
and NADPH would not be produced. Proton translocation from
the stroma to the thylakoid lumen also would not occur, so CF1
would not be stimulated and ATP synthesis would not occur.
28. Because oligomycin inhibits the F0 subunit of mitochondrial ATP synthase, protons from the intermembrane space cannot
translocate to the mitochondrial matrix. F1 is not stimulated, and
ATP synthesis does not occur. The cytosolic ratio of ATP/ADP decreases because less ATP is exported to the cytosol from the matrix
when ATP synthesis is inhibited. Because oligomycin does not inhibit CF0, ATP synthesis is not inhibited, so the ATP/ADP ratio is
not affected. In some cases the decreased cytosolic ATP/ADP ratio
might stimulate ATP synthesis in the chloroplast, resulting in an
increased chloroplastic ATP/ADP ratio. [From Gardeström, P., and
Lernmark, U., J. Bioenerg. Biomembr. 27, 415–421 (1995).]
30. The net equation would be
CO2 1 2 H2S S (CH2O) 1 S2 1 H2O
32. When rubisco reacts with oxygen in a process called
photorespiration, the products are 3-phosphoglycerate and
2-phosphoglycolate. The latter product is further metabolized
by pathways that consume ATP and NADPH and thus waste
some of the energy captured by photosynthesis. If rubisco could
bind only CO2, the enzyme would produce two molecules of
3-phosphoglycerate, which is a precursor to many biological
macromolecules. In a plant with a rubisco that binds only CO2,
photorespiration would not occur, and more of the energy obtained
in photosynthesis would be available for biosynthetic reactions.
34.
DG 5 DG °¿ 1 RT ln Q
241.0 kJ ? mol21 5 235.1 kJ ? mol21 1 18.3145
3 1023 kJ ? K21 ? mol21 2 1298 K2 lnQ
25.9 kJ ? mol21 5 2.48 kJ ? mol21 ln Q
22.38 5 ln Q
e22.38 5 Q
Q 5 0.093
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enters glycolysis to produce ATP to meet the energy needs of
the plant. [From Cséke, C., Nishizawa, A. N., and Buchanan,
B. B., Plant Physiol. 70, 658–661 (1982).]
40.
Carbon
fixation
Light
P700*red
Ferredoxinox
38. ATP and NADPH are produced by the light reactions
during the day. When photosynthesis is occurring, synthesis
of starch (an ATP-requiring process) occurs, and glycolysis,
an opposing process, is inhibited. At night, when the light
reactions do not occur and ATP and NADPH concentrations
are low, the inhibition of phosphofructokinase is relieved.
Starch is broken down, and the resulting glucose-1-phosphate
40
| CHAPTER 17 Solutions
Thioredoxinox
SH SH
P700⫹ox
Ferredoxinred
S
Reductaseox
S
S
Thioredoxinred
SH SH
S
Enzymered
SH SH
Enzymeox
S
S
The activity of Photosystem I generates reduced ferredoxin, which
is a substrate for the reductase. The product of the reductase reaction, reduced thioredoxin, can then reduce the disulfides of the Calvin cycle enzymes. Conformational changes in the enzymes upon
exposure of free sulfhydryl groups could increase their activity.
42.
DG 5 DG °¿ 1 RT lnQ
229.7 kJ ? mol21 5 214.2 kJ ? mol21 1 18.3145
3 1023 kJ ? K21 ? mol21 2 1298 K2 lnQ
215.5 kJ ? mol21 5 2.48 kJ ? mol21 lnQ
26.25 5 lnQ
e26.25 5 Q
Q 5 0.0019
The enzyme is likely to be regulated because it catalyzes an irreversible step of the Calvin cycle (as evidenced by the large negative value of DG).
44. PEPC supplies oxaloacetate for the citric acid cycle, which
can generate ATP for the biosynthetic processes occurring in
the germinating seedling. The PEPC enzyme also plays a role in
providing oxaloacetate for the glyoxylate cycle (see Box 14-B),
which provides a route for the synthesis of glucose from fatty
acids, a pathway that is lacking in animals.
Chapter 17 Solutions
O
2.
R
C
O⫺
The ratio of products to reactants is 0.093 to 1.
36. (a) The compound resembles the transition state of the
rubisco carboxylase reaction (see Fig. 16-24) and therefore
inhibits the enzyme by binding in the active site.
(b) At night, the compound inhibits rubisco in order to
prevent the Calvin cycle from consuming NADPH and
ATP. During the day, when the light reactions are supplying
NADPH and ATP, the inhibitor is broken down to reactivate rubisco so that it can fix carbon.
Reductasered
Fatty Acid
⫹
O
⫺O
O
P
O
O⫺
O
P
O
O⫺
P
O
Adenosine
O⫺
ATP
inorganic
pyrophosphatase
PPi
2 Pi
H 2O
O
R
C
O
O
P
O
Adenosine
O⫺
H
SCoA
O
Adenosine
Acyladenylate
mixed anhydride
O
R
C
O
SCoA
Acyl-CoA
⫹
⫺O
P
O⫺
AMP
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4. If medium-chain fatty acids are metabolized normally in the
absence of carnitine, then fatty acids of this size must be able to
cross the inner mitochondrial membrane and enter the mitochondrial matrix without the aid of carnitine. Carnitine is required for transport of fatty acids long than 10 carbons.
6. The patient should avoid fasting and sustained exercise, since
both of these conditions result in the release of free fatty acids into
the bloodstream for uptake by the tissues. The patient should also
consume a diet that is low in fat and high in carbohydrates.
8. Oxidation of the methylene groups is accomplished by the
action of a dehydrogenase, which removes electrons. The oxygen
atom that becomes part of the carbonyl group is derived from a
water molecule added by enoyl-CoA hydratase.
10. The products are three molecules of propionyl-CoA, three
molecules of acetyl-CoA, and one molecule of 2-methylpropionylCoA.
O
O
␤ oxidation
H3 C
CH2
C
S
CoA
Propionyl-CoA
C SCoA
(CH2)7
CH
O
(CH2)7
CH
C
SCoA
Oleoyl-CoA
3 cycles of ␤ oxidation
H3C
(CH2)7
3 (1 QH2, 1 NADH, 1 acetyl-CoA)
3(1.5 2.5 10) 42 ATP
O
C
C
H
H
CH2
C
SCoA
enoyl-CoA isomerase
H3C
(CH2)8
C
H
O
C
C
SCoA
H
1 cycle of ␤ oxidation
(acyl-CoA dehydrogenase
reaction is bypassed)
1 NADH, 1 acetyl-CoA 2.5 10 12.5 ATP
O
H3C
(CH2)8
C
SCoA
4 cycles of ␤ oxidation
4 (1 QH2, 1 NADH, 1 acetyl-CoA) 4(1.5 2.5 10) 56 ATP
Acetyl-CoA (10 ATP)
C SCoA
Pristanate
O
H3C
(b) Linoleate is also an 18-carbon fatty acid and would yield
120 ATP via b oxidation if it were saturated. The double
bond at the 9,10 must be converted from the cis to the trans
isomer, resulting in the loss of 1 QH2 (1.5 ATP). In addition, an NADPH-dependent reduction of the 12,13 double
bond costs 2.5 ATP. This reduces the ATP total by 4, so
oxidation of linoleate yields 116 ATP. (Subtract 2 ATP for
activation from the total shown in the figure.)
O
␤ oxidation
C
CH3
SCoA
Acetyl-CoA
O
H3C
(CH2)4
CH
CH
␤ oxidation
H3C
SCoA
(CH2)4
CH
CH
CH2
C
C
H
H
CH2
C
H
O
C
C
CoA
enoyl-CoA isomerase
H3C
(CH2)4
CH
CH
CH2
CH2
C
CoA
H
Acetyl-CoA
␤ oxidation
C
O
O
Propionyl-CoA
␤ oxidation
(CH2)7
3 (1 QH2, 1 NADH, 1 acetyl-CoA) 3(1.5 2.5 10) 42 ATP
3 cycles of ␤ oxidation
CH
Linoleyl-CoA
SCoA
CH
CH2
1 cycle of ␤ oxidation
(acyl-CoA dehydrogenase
reaction is bypassed)
1 NADH, 1 acetyl-CoA 2.5 10 12.5 ATP
O
Propionyl-CoA
H3C
(CH2)4
CH
CH
CH2
CH2
C
S
CoA
acyl-CoA dehydrogenase 1 QH2 1.5 ATP
␤ oxidation
Acetyl-CoA
H3C
CH3
SCoA
H3C
(CH2)4
CH
CH
C
H
O
C
C
H
NADPH
NADP
2,4-dienoyl-CoA reductase
CoA
Costs 2.5 ATP
O
12. (a) Oleate is an 18-carbon fatty acid. If it were completely
saturated, 120 ATP would be generated via b oxidation
(see Problem 11b). The double bond at the 9,10 position
must be converted from the cis to the trans isomer. b oxidation continues after this isomerization step, but with the
loss of 1 QH2. Thus 1.5 ATP should be subtracted from
the total to yield 118.5 ATP. The pathway below shows a
total of 120.5 ATP, but 2 ATP are subtracted to account for
the ATP spent in fatty acid activation.
S
O
H3C
(CH2)4
CH2
CH
CH
CH2
C
S
CoA
S
CoA
enoyl-CoA isomerase
H3C
(CH2)4
CH2
CH2
H
O
C
C
C
H
1 cycle of ␤ oxidation
(acyl CoA dehydrogenase
reaction is bypassed)
3 cycles of ␤ oxidation
1 NADH, 1 acetyl-CoA (2.5 10) 12.5 ATP
3 (1 QH2, 1 NADH, 1 acetyl-CoA) 3(1.5 2.5 10) 42 ATP
Acetyl-CoA (10 ATP)
CHAPTER 17 Solutions | 41
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14. Adding up the ATP in the figure gives a total of 155 ATP.
Subtracting the 2 ATP required for activation brings the total to
153 ATP.
O
H3C
(CH2)20
CH2
acyl-CoA oxidase
H3C
(CH2)20
enoyl-CoA
C
CH2
H
O
C
C
SCoA
2 H2O O2
H2O2
O2
SCoA
H
6 cycles of ␤ oxidation
(acyl-CoA dehydrogenase
reaction is bypassed)
H3C
C
FAD
FADH2
(CH2)8
6 NADH, 6 acetyl-CoA 6(2.5) 6(10) ATP 75 ATP
O
CH2
CH2
C
SCoA
PEROXISOME
MITOCHONDRION
5 QH2, 5 NADH, 5 acetyl-CoA 5(1.5) 5(2.5) 5(10) ATP 70 ATP
acetyl-CoA (10 ATP)
5 cycles of ␤ oxidation
16. (a) The free energy cost of synthesizing ATP from ADP 1 Pi
is 30.5 kJ ? mol21. Glucose oxidation could theoretically
yield 2850/30.5, or 93.4 ATP. The ATP yield per carbon
atom is 93.4/6 5 15.6. Palmitate oxidation could theoretically yield 9781/30.5, or 320.7 ATP. The ATP yield per carbon atom is 320.7/16 5 20.4 ATP.
(b) In vivo, the catabolism of glucose leads to the production
of 32 ATP (see Solution 15-31), which is 32/6, or 5.3 ATP
per carbon atom. The catabolism of palmitate leads to the
production of 106 ATP (see the Solution 17-11a), which is
106/16, or 6.6 ATP per carbon atom.
(c) In theory as well as in vivo, fatty acid oxidation yields
more ATP per carbon atom than glucose oxidation (this is
primarily because the carbons of carbohydrates are already
partially oxidized, whereas fatty acid carbons are usually
fully reduced). Both pathways are equally efficient from
a thermodynamic point of view, recovering slightly more
than 30% of the free energy available (for glucose 5.3/15.6,
and for palmitate, 6.6/20.4).
18. Acetyl-CoA carboxylase generates malonyl-CoA, the substrate for fatty acid synthase. Mice without acetyl-CoA carboxylase are unable to synthesize fatty acids and hence store less fat in
their bodies. Malonyl-CoA also inhibits carnitine acyltransferase.
In the absence of acetyl-CoA carboxylase to produce this inhibitor, transport of fatty acids into mitochondria cannot be regulated, and mitochondrial b oxidation proceeds continuously.
20. As shown in the mechanism for malonyl-CoA formation
(Problem 17-19), ATP is required to generate a carboxyphosphate intermediate. The result is that malonyl-CoA contains
some of the free energy of ATP. This free energy is released when
the acetyl group condenses with malonyl-ACP and CO2 is released. The release of CO2 is entropically favored and drives the
reaction to completion.
42
| CHAPTER 17 Solutions
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22. Epinephrine signaling via an adrenergic receptor activates a G protein, which activates adenylate cyclase to produce cyclic AMP to activate protein kinase A (PKA). PKA
phosphorylates its substrates, including acetyl-CoA carboxylase, thereby inactivating the enzyme. As a result of lower
acetyl-CoA carboxylase activity, the rate of fatty acid synthesis drops. Epinephrine signaling also leads to phosphorylation
and activation of glycogen phosphorylase, which mobilizes
glucose from glycogen. These responses are consistent: When
the cell needs to mobilize metabolic fuel (for example, by
glycogenolysis), storage of fuel (for example, by fatty acid
synthesis) is inhibited.
24. The label does not appear in palmitate because 14CO2 is
released in Reaction 3 of fatty acid synthesis (Fig. 17-13).
26. (a) A portion of the inhibitor molecule mimics the structure
of a fatty acid. The compound may act as a competitive inhibitor, binding to the enzyme active site to preclude binding of the substrate.
(b) The lower the ID50 value, the lower the concentration
of inhibitor required to kill the cells. It is desirable for the
inhibitor to inhibit fatty acid synthase in cancer cells but
not normal cells. Thus, an effective inhibitor would have
a low ID50 value in cancer cells but a high ID50 value in
normal cells. The ratios of ID50 values for normal and cancer cells were calculated to determine which of the inhibitors tested was most effective. As shown in the table below,
compounds B, D, and E were the most effective inhibitors.
These compounds have side chains of 11, 7, and 8 carbons,
respectively. Inhibitor D has the shortest alkyl chain and is
the most soluble of the three.
Compound
A
B
C
D
E
F
Alkyl side chain (R)
—C13H27
—C11H23
—C9H19
—C8H17
—C7H15
—C6H13
breast cancer ID50
normal cells ID50
2.7
6.0
2.5
4.3
4.5
1.5
(c) Inhibitor D at a concentration of 2 mg/mL significantly inhibits incorporation of radiolabeled acetate into
triacylglycerols. At a concentration of 5 mg/mL, 80% of
total acylglyceride synthesis is inhibited. Phospholipid synthesis is not significantly affected, supporting the hypothesis
that the drug’s target is fatty acid synthase. [From Kuhajda,
F. P., Pizer, E. S., Li, J. N., Mani, N. S., Frehywot, G. L.,
and Townsend, C., Proc. Natl. Acad. Sci. 97, 3450–3454
(2000).]
28. DHA is an n-3 fatty acid and is essential for growth. A wellnourished mother will produce breast milk containing DHA,
but a bottle-fed infant whose sole source of food is infant formula will not obtain this essential fatty acid unless it is added to
the formula.
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30. (a) Liver fatty acid synthase activity increases with consumption of a high-carbohydrate diet. Glucose in excess of
what is required to meet immediate energy needs is oxidized
to pyruvate by glycolysis, then converted to acetyl-CoA by
pyruvate dehydrogenase. Excess acetyl-CoA is used to synthesize fatty acids, which are ultimately used to synthesize
triacylglycerols for storage in adipose tissue.
(b) Liver fatty acid synthase activity is decreased with consumption of a high-fat diet. Endogenous fatty acid synthesis
is not required if fatty acids are being obtained from the
diet.
(c) Mammary gland fatty acid synthase activity increases in
mid to late pregnancy to provide fatty acids for triacylglycerols in breast milk for the neonate.
32. Fatty acid degradation produces acetyl-CoA, but in the
absence of glucose, acetyl-CoA cannot enter the citric acid
cycle due to insufficient oxaloacetate, a substrate for the first
step. When carbohydrate concentrations drop, any available
oxaloacetate is diverted to gluconeogenesis to produce the glucose necessary for red blood cell and brain function. If acetylCoA cannot enter the citric acid cycle (and acetyl-CoA cannot be converted to pyruvate in mammals), it is converted to
ketone bodies.
34. Oxidation of fatty acids yields acetyl-CoA, which enters
the citric acid cycle by reacting with oxaloacetate. Oxaloacetate is produced by the action of pyruvate carboxylase on
pyruvate produced by glycolysis. Without glucose metabolism
to generate pyruvate, the citric acid cycle cannot proceed.
Thus, the statement “fats burn in the flame of carbohydrates”
refers to the fact that normal carbohydrate metabolism (leading production of oxaloacetate) is absolutely required for normal fatty acid metabolism, which provides acetyl-CoA for the
citric acid cycle.
36. Three phosphoanhydride bonds must be cleaved in order to
synthesize phosphatidylcholine from choline and diacylglycerol
(see Fig. 17-19). Choline is activated by conversion to phosphocholine, with a phosphate group donated by ATP. The phosphocholine is subsequently converted to CDP–choline, a reaction
in which a second phosphoanhydride bond is cleaved. Pyrophosphate is the leaving group in this reaction, and the third
phosphoanhydride bond is cleaved when the pyrophosphate is
hydrolyzed to orthophosphate.
38. (a) HDL remove excess cholesterol from tissues and transport it back to the liver. This prevents the accumulation of
cholesterol in vessel walls that leads to atherosclerosis.
(b) HDL level alone does not indicate the risk of developing
atherosclerosis, since the level of LDL, the activity of the
LDL receptor, and other factors such as smoking or vessel
wall injuries resulting from infection can all influence the
likelihood of developing the disease.
40. Chylomicrons are produced by intestinal cells to package
dietary lipids for delivery to the rest of the body. Fat-soluble
vitamins such as vitamin A are also transported via chylomicrons. Without chylomicrons, the body lacks an efficient way to
distribute vitamin A obtained from the diet.
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Chapter 18 Solutions
2. Alfalfa is a leguminous plant (see Fig. 18-1) that harbors
nitrogen-fixing bacteria in its root nodules. Crop growing depletes the soil of fixed nitrogen; planting alfalfa every few years
allows the associated nitrogen-fixing bacteria to replenish the
nitrogen supply in the soil and decreases the need for nitrogenbased fertilizers.
4. Bacterial nitrogenase is inactivated by oxygen, so nitrogen
fixation takes place in cells that lack Photosystem II, the complex where water is oxidized to form molecular oxygen (see
Section 16-2)
6. (a) The low K M of glutamine synthetase for ammonium ions
ensures that the first method (a) will be used when the concentration of available ammonia is low. The second method
(b) will be used when the concentration of ammonia is high.
(b) Incorporation of one mole of ammonia into an amino acid
using the first method (a) costs one mole of ATP, whereas ATP is
not required when the prokaryotic cell uses the second method
(b). The prokaryotic cell is at a disadvantage when the ammonia concentration is low because energy must be expended in
order to synthesize amino acids under these conditions.
8.
COO
(a)
O
Glu
H
3N
CH
Asp
3N
Ala
Glu
O
C
3N
COO
Asp
O
C
Oxaloacetate
CH2
Pyruvate
CH3
O
C
O
C
-Ketoglutarate
O
CH
Oxaloacetate
COO
Glu
CH3
(c)
H
CH2
COO
O
CH
O
C
-Ketoglutarate
COO
(b)
H
O
C
CH2
CH2
-Ketoglutarate
CH2
CH2
COO
COO
The products are all intermediates of the citric acid cycle (oxaloacetate and a-ketoglutarate) or closely associated with the citric
acid cycle (pyruvate).
10.
E
(CH2)4
NH2
H3C
H
O
H
C
H
C
H
C
COO
N
H
O
O32PO
N
H
O
H3C
E
H
C H
COO
C
H
NH
3
(CH2)4
N
H
C
O
O32PO
N
H
12. In the glutamine synthetase reaction (shown on page 468),
ATP donates a phosphoryl group to glutamate, which is then
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displaced by an ammonium ion, producing glutamine and
phosphate. The ammonium ion is the nitrogen source. The asparagine synthetase reaction (page 472) also requires ATP as an
energy source, but the nitrogen donor is glutamine, not an ammonium ion. Aspartate is converted to asparagine, and the glutamine becomes glutamate after donating an amino group. ATP
is hydrolyzed to AMP and pyrophosphate instead of ADP and
phosphate as in the glutamine synthetase reaction.
14. Cysteine is the source of taurine. The cysteine sulfhydryl
group is oxidized to a sulfonic acid, and the amino acid is decarboxylated.
16. Because the genetically modified soybeans have the bacterial
EPSPS enzyme, the plants are resistant to the Roundup® herbicide. These plants will be able to synthesize chorismate (and aromatic amino acids) even in the presence of the herbicide. Weeds
will not be resistant, so the herbicide will kill the weeds but not
the soybean plant.
18. (a) Collagen contains a high proportion of Gly and Pro residues and relatively few of the other amino acids (Section 5-2).
Therefore, it does not provide a good mix of amino acids.
(b) Because collagen is a protein, it supplies amino acids
that can be used as metabolic fuels or converted to carbohydrates, lipids, or nucleotides. Sucrose (glucose and fructose)
is a metabolic fuel, but its carbons cannot be used to synthesize amino acids or nucleotides unless a source of amino
groups is also provided.
20. (a) Lysine might regulate its own synthesis by acting as an
allosteric inhibitor of dihydropicolinate synthase. This enzyme catalyzes the first unique reaction in a series of eight
reactions that lead to lysine.
(b) Using the same reasoning, methionine might acts as a
feedback inhibitor for acyltransferase. Methionine might
also regulate the activity of homoserine dehydrogenase, another major branch point of the pathway, even though this
enzyme is also involved in threonine biosynthesis.
(c) Similarly, threonine might act as a feedback inhibitor
for homoserine kinase and might also inhibit homoserine
dehydrogenase.
22. (a) A 10-fold increase in the K M for APRT results in decreased affinity of the enzyme for one of its substrates. The
result is the accumulation of adenine, which is oxidized to
dihydroxyadenine and forms kidney stones.
(b) One way to treat this condition is to administer a compound that would bind to xanthine dehydrogenase and prevent adenine from binding. This would prevent the conversion of adenine to dihydroxyadenine.
24. Phosphoribosyl pyrophosphate (PRPP) is a reactant in the
salvage reactions involving IMP and GMP, so if these reactions
cannot occur, the PRPP that would normally be used in the salvage reactions would accumulate. PRPP stimulates amidophosphoribosyltransferase by feed-forward activation (see Problem 21),
which accelerates the synthesis of purine nucleotides and increases
the concentration of their degradation product, uric acid.
26. 5-Fluorouracil structurally resembles uracil and competitively
inhibits thymidylate synthase. When the enzyme is inhibited,
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| CHAPTER 18 Solutions
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dUMP is not converted to dTMP (see page 484) and there is
no subsequent phosphorylation to dTTP, so the concentration
of dTTP in the cells declines. In the presence of the inhibitor,
dUMP accumulates and is converted to dUTP. The concentration of dUTP in cells is normally low because thymidylate synthase acts quickly on dUMP to convert it to dTMP. Cancer cells
die in the presence of 5-fluorouracil because the cells lack the
dTTP required for DNA synthesis. Normal cells are not affected
because DNA synthesis is not as rapid as in cancer cells.
28. The ketogenic amino acids are broken down to acetyl-CoA
or acetoacetate, which can be converted to acetyl-CoA. For glucogenic amino acids, the carbon skeletons are broken down to
either pyruvate (which is converted to acetyl-CoA by the pyruvate dehydrogenase complex) or a citric acid cycle intermediate
(which can be converted to phosphoenolpyruvate by the reactions of gluconeogenesis; phosphoenolpyruvate is the precursor
of pyruvate and therefore of acetyl-CoA).
30. The fate of propionyl-CoA produced upon degradation of
isoleucine is identical to that of propionyl-CoA produced in the
oxidation of odd-chain fatty acids (see Fig. 17-7). Propionyl-CoA
is converted to (S )-methylmalonyl-CoA by propionyl-CoA carboxylase. A racemase converts the (S )-methylmalonyl-CoA to the
(R) form. A mutase enzyme converts the (R )-methylmalonyl-CoA
to succinyl-CoA, which enters the citric acid cycle.
32. The missing enzyme is the branched chain a-keto acid
dehydrogenase (see Fig. 18-11). This enzyme is common to
the catabolic pathways of valine, isoleucine (see Problem 29),
and leucine (see Problem 31) and explains why the corresponding a-keto acids of these amino acids accumulate. Because the patients are unable to break down branched-chain
amino acids, they should consume a diet that contains only
enough of these amino acids to meet the needs of growth and
development.
34. (a)
O
CH2
C
COO
Phenylpyruvate
(b) A tetrahydrobiopterin deficiency prevents the conversion of phenylalanine to tyrosine in the phenylalanine catabolic pathway. Consequently, phenylalanine accumulates
and undergoes transamination to phenylpyruvate, which is
excreted.
(c) The growing child still requires some phenylalanine for
growth. The low-phenylalanine diet should provide enough
phenylalanine for growth but should not exceed amounts
needed for growth since the excess cannot be metabolized
due to the lack of the PAH enzyme.
(d) The artificial sweetener aspartame consists of a methylated Asp–Phe dipeptide. (The C-terminal carboxyl group is
methylated.) Aspartame is broken down to Asp, Phe, and
methanol. Since aspartame is a source of phenylalanine,
patients with PKU should not use this product, and physicians should advise their patients to check labels carefully
and to avoid use of this product.
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(e) Phenylalanine is the precursor of tyrosine. If the phenylalanine intake is low, supplemental tyrosine may be needed.
36. Proteins, a polymeric form of amino acids, could be considered as a storage depot for amino acids, since the proteins can
be degraded to release amino acids for use as metabolic fuels.
However, proteins have functions other than fuel storage, which
is not the case for glycogen and triacylglycerols (although triacylglycerols also function as thermal insulation in some species).
38. Glutamate dehydrogenase, glutamine synthetase, and carbamoyl phosphate synthetase can potentially “mop up” NH14.
40. Adding arginine, the product of the argininosuccinase reaction, would increase flux through the urea cycle.
42. An individual consuming a high-protein diet uses amino
acids as metabolic fuels. As the amino acid skeletons are converted to glucogenic or ketogenic compounds, the amino groups
are disposed of as urea, leading to increased flux through the
urea cycle. During starvation, proteins (primarily from muscle)
are degraded to provide precursors for gluconeogenesis. Nitrogen from these protein-derived amino acids must be eliminated,
which demands a high level of urea cycle activity.
44. (a) GTP (representing a high level of cellular ATP) inhibits
glutamate dehydrogenase to favor glutamate formation.
(b) ADP (a signal for low cellular energy) activates glutamate dehydrogenase to favor synthesis of a-ketoglutarate so
that citric acid cycle flux can increase.
(c) NADH inhibits glutamate dehydrogenase to favor glutamate formation, since a high level of NADH means that
the cell does not need the citric acid cycle to produce additional reduced cofactors.
46. (a) H. pylori urease converts urea to NH3 and CO2 (see
page 494). The ammonia has a pK of 9.25, so it combines
with protons to produce NH14. The resulting decrease in
hydrogen ion concentration helps the bacteria maintain an
intracellular pH higher than the environmental pH.
(b) Urease on the cell surface increases the pH of the fluid
surrounding the cell, creating a more hospitable microenvironment for bacterial growth.
Chapter 19 Solutions
2. Glucose-6-phosphate is the product of the hexokinase reaction and is produced when glucose is transported into cells. G6P
can then continue along the pathway of glycolysis to produce
ATP and biosynthetic intermediates. G6P can be reversibly converted to glucose-1-phosphate for incorporation into glycogen.
Glycogen degradation produces glucose-1-phosphate, which is
isomerized to glucose-6-phosphate. G6P is also the product of
gluconeogenesis. Glucose-6-phosphatase catalyzes the removal
of the phosphate group to produce glucose, which can leave the
cell. Glucose-6-phosphate can also enter the pentose phosphate
pathway to produce NADPH and ribose.
4. Glucose enters the red blood cell via specific transport
proteins and undergoes glycolysis to produce pyruvate. The
next step is the conversion of pyruvate to lactate with concomitant production of NAD1. The NAD1 is used in the
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glyceraldehyde-3-phosphate dehydrogenase reaction, and its
production is essential for the continuation of the glycolytic
pathway. Despite the presence of oxygen in the red blood cell,
glucose is oxidized anaerobically due to the lack of mitochondria. Two ATP are produced per glucose molecule.
6. (a) The reaction is probably a near-equilibrium reaction because the reactants and products have the same total number
of phosphoanhydride bonds.
(b) In highly active muscle, ATP is rapidly converted to ADP.
Adenylate kinase catalyzes the conversion of two ADP to
ATP and AMP as a way to generate additional ATP to power
the actin–myosin contractile mechanism.
8. When AMP concentrations are high, the energy status of the
cell is low, and the glycolytic pathway is needed to provide ATP
for the cell. AMP deaminase acts on the increased concentrations
of AMP to produce ammonium ions. These, along with AMP,
stimulate the activities of phosphofructokinase and pyruvate kinase and increase the activity of the glycolytic pathway as a whole.
Factors that influence the activity of adenosine deaminase will
thus also have an indirect effect on the rate of glycolysis. [From
Yoshino, M., and Murakami, K., J. Biol. Chem. 260, 4729–4732
(1985).]
10. The lactate dehydrogenase reaction, which reduces pyruvate to lactate, occurs with concomitant oxidation of NADH
to NAD1. NAD1 serves as a reactant in the glyceraldehyde-3phosphate reaction in glycolysis. If pyruvate were the end product of glycolysis, all of the cellular NAD1 would become reduced to NADH and the glycolytic pathway would grind to a
halt for lack of NAD1.
12. Plasma alanine would be elevated as well. The increased
plasma concentrations of pyruvate would accelerate the operation of the glucose–alanine cycle. Pyruvate is taken up by the
muscle and transaminated to alanine, then released to the circulation to be taken up by the liver. Since plasma pyruvate levels
are elevated, alanine levels would also be elevated.
14. Increased levels of citrulline indicate that the cytosolic argininosuccinate synthetase reaction in the urea cycle (see Section
18-5) is not occurring normally. This reaction requires aspartate as a reactant in addition to citrulline. The accumulation of
citrulline may occur because there is a shortage of aspartate. A
deficiency of pyruvate carboxylase results in a decreased concentration of oxaloacetate that could otherwise be transaminated to
aspartate. Hyperammonemia is the result of urea cycle impairment, since ammonia is not being converted to urea for excretion by the kidneys. [From Coude, F. X., Ogier, H., Marsac, C.,
Munnich, A., Charpentier, C., and Saudubray, J. M., Pediatrics
68, 914 (1981).]
16. Because glucokinase is not saturated at physiological
glucose concentrations, it can respond to changes in glucose
availability with an increase or decrease in reaction velocity.
Consequently, the entry of glucose into glycolysis and subsequent metabolic pathways depends on the glucose concentration. Because hexokinase is saturated at physiological glucose
concentrations, its rate does not change with changes in glucose concentration.
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18. Insulin stimulates uptake of glucose via GLUT4 receptors
in adipocytes. Glucose is converted to glycerol-3-phosphate via
the enzymes of the glycolytic pathway. The glycerol-3-phosphate
is required for the backbone of the triacylglycerol molecule.
20. Ingesting large amounts of glucose stimulates the b cells
of the pancreas to release insulin, which causes liver and muscle
cells to use the glucose to synthesize glycogen and causes adipose
tissue to synthesize fatty acids. Insulin also inhibits the breakdown of metabolic fuels. The body is in a state of resting and
digestion and is not prepared for running.
22. Phosphorylation of glycogen synthase by GSK3 inactivates
the enzyme so that glycogen synthesis does not occur. But when
insulin activates protein kinase B, the kinase phosphorylates
GSK3. Phosphorylated GSK3 is inactive and unable to phosphorylate glycogen synthase. In the dephosphorylated state glycogen synthase is active and glycogen synthesis can occur.
24. (a) Normally, glucagon binds to cell-surface receptors on the
liver, stimulating adenylate cyclase to produce cAMP and
activate protein kinase A, which subsequently activates glycogen phosphorylase via phosphorylation. Glycogen phosphorylase catalyzes the degradation of glycogen to glucose,
which is released into the bloodstream. Blood glucose concentrations should rise shortly after an intravenous injection
of glucagon. Glycogen degradation in the patient’s liver thus
appears to be normal.
(b) Glycogen metabolism in the liver appears to be normal,
since glycogen content is normal and the patient’s response
to the glucagon test is normal. However, muscle glycogen is
elevated, which indicates a defect in muscle glycogen metabolism. Most likely glycogen synthesis is normal, since
the biochemical structure of the glycogen is normal. A deficiency in the muscle glycogen phosphorylase enzyme is the
most likely explanation.
(c) Blood alanine concentrations normally increase as part
of the glucose–alanine cycle. Glucose-6-phosphate, the
product of glycogenolysis, enters glycolysis and produces
pyruvate. Pyruvate undergoes a transamination reaction
to form alanine, which is released from the muscle. Alanine then enters the liver, where the transamination reaction takes place in reverse to re-form pyruvate. Pyruvate can
then enter gluconeogenesis, the resulting glucose product is
released, and the cycle begins again. But the patient’s tissues
cannot perform the glucose–alanine cycle because his muscles are unable to produce glucose-6-phosphate. Instead, the
muscles take up alanine and use it as a fuel. Thus, plasma
alanine levels in the patient decrease rather than increase.
(d) Blood glucose concentrations are regulated by pancreatic
hormones acting on the liver to stimulate glycogen synthesis
or degradation, whatever is appropriate. Since the patient’s
liver enzymes appear to function normally, his blood glucose concentration is properly regulated and he is neither
hypo- nor hyperglycemic.
(e) If the patient ingests glucose or fructose, blood sugar
concentrations can be maintained at a high level. The muscles will thus be able to take up glucose or fructose and oxidize these sugars via glycolysis to yield the ATP required of
46
| CHAPTER 19 Solutions
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active muscles. In this way, the muscles do not have to rely
on stored glycogen as a fuel source.
26. (a) PAH is an oligomer with a subunit size of about 49,000
daltons. The subunits dissociate upon treatment with SDS,
so one band at the low molecular weight is observed. The
subunits are identical, since there is a single band in this
lane. If PAH is subjected to electrophoresis under nondenaturing conditions, a single band with a molecular weight
slightly less than 100,000 daltons is observed. Since this is
double the molecular weight of the monomer, it can be concluded that the PAH consists of a dimer under nondenaturing conditions. Preincubation of PAH with phenylalanine
results in a band that is four times the molecular weight of
the monomer. Therefore, PAH is converted from a dimer to
a tetramer in the presence of phenylalanine.
(b) PAH is a dimer that can be converted to a tetramer at
high concentrations of phenylalanine. The sigmoidal curve
indicates that the binding of the substrate is cooperative:
The binding of one substrate molecule facilitates the binding of subsequent substrate molecules. Thus, PAH is an
allosteric enzyme and likely is affected by allosteric modulators that stimulate or inhibit the enzyme and thus control
the degradation rate of phenylalanine.
(c) As discussed above, preincubation with phenylalanine
converts PAH from the dimer to a tetrameric form. It is
possible that the tetrameric form has a greater affinity for
substrate than the dimeric form, which would explain the
increase in velocity when the enzyme is preincubated with
Phe. The dimer is the T form, which has a low affinity for
the substrate, and the tetramer is the R form, which has a
high affinity for the substrate.
(d) Insulin inhibits the activity of PAH, whereas glucagon
stimulates the activity of the enzyme. Because PAH is phosphorylated in the presence of glucagon, it’s likely that glucagon exerts its effects by activating protein kinase A, which
in turn phosphorylates the enzyme. The effects of glucagon
and Phe appear to be synergistic, since the enzyme is more
active when glucagon and Phe are combined than when either stimulant is used alone. Since glucagon concentrations
are high when blood sugar is low, it’s possible that PAH
activation occurs in these circumstances to stimulate the
phenylalanine degradation pathway to produce fumarate,
which can enter gluconeogenesis. [From DiLella, A. G.,
Marvit, J., and Woo, S. L. C., (1987) The Molecular Genetics of Phenylketonuria in Amino Acids in Health and Disease:
New Perspectives, Alan R. Liss, Inc., pp. 553–564.]
28. Several days into a fast, muscle and liver glycogen have been
depleted. In the absence of dietary glucose, the main source of
endogenous glucose is gluconeogenesis. Citric acid cycle intermediates are used to synthesize oxaloacetate and then pyruvate,
which enters the gluconeogenic pathway. Catalytic amounts of
citric acid cycle intermediates are required for proper functioning of the cycle, so when these intermediates are diverted to gluconeogenesis, the citric acid cycle cannot function properly.
30. Regular exercise maintains muscle mass so that muscle proteins are less likely to be broken down to provide amino acids
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as fuel, and the body uses its stored fat instead. Inactivity while
dieting may promote loss of muscle mass instead of fat.
32. (a) The cytochrome c content is high because of the large
number of mitochondria that allow brown adipose tissue to
oxidize metabolic fuels aerobically, funneling the reduced
coenzymes through the electron transport chain. Uncoupling oxidative phosphorylation permits the energy of electron transport to be dissipated as heat (see Box 15-A).
(b) When subjects were exposed to cold, uptake of the labeled
glucose into brown adipose tissue increased to provide reduced
coenzymes for electron transport and thermogenesis.
34. Endogenous fatty acid synthesis is required when dietary
fatty acid intake is insufficient. Stimulation of acetyl-CoA carboxylase ensures that the cell will have enough fatty acids in
the absence of dietary lipids (although essential fatty acids will
still be lacking under these circumstances). During starvation
(and untreated diabetes, which is similar to the starved state),
body tissues do not have the resources to synthesize fatty acids,
so acetyl-CoA carboxylase enzymes are inhibited. In the starved
state, fatty acids are mobilized to provide fuel to body tissues.
36. (a) Acetate is converted to acetyl-CoA, which serves as a
substrate for acetyl-CoA carboxylase and is converted to
malonyl-CoA. Fatty acid synthase normally acts on malonylCoA for incorporation into fatty acids, but in the presence
of the inhibitor C75, fatty acid synthesis does not occur.
Therefore, radioactive malonate accumulates, and virtually
no label appears in fatty acids.
(b) C75 causes hypothalamic levels of NPY to decrease.
This decreases appetite, which leads to weight loss. In this
manner, C75 fools the body into thinking that it is in the
fed state, when in fact it is being starved.
(c) Acetyl-CoA carboxylase (ACC) catalyzes the reaction
that converts acetyl-CoA to malonyl-CoA. In the presence
of the ACC inhibitor, malonyl-CoA cannot be produced
and hepatic malonyl-CoA levels would decrease. Subsequent administration of C75 would have no effect on the
ACC inhibitor–treated animals, because C75, a fatty acid
synthase inhibitor, acts downstream of ACC. If malonylCoA inhibits feeding, the ACC inhibitor–treated and C75treated mice would have lower levels of malonyl-CoA, and
feeding would not be inhibited, that is, the mice would eat
normally. These experiments were carried out and this is in
fact what occurred.
(d) When malonyl-CoA accumulates, long-chain fatty acylCoA molecules also accumulate in the cytosol. Malonyl-CoA
inhibits carnitine acyltransferase, which prevents translocation of fatty acyl-CoAs into the mitochondrial matrix for
b oxidation. It is possible that long-chain fatty acyl-CoAs
serve as signaling molecules to begin the pathway that suppresses appetite. [From Loftus, T. M., Jaworksy, D. E., Grehywot, G. L., Townsend, C. A., Ronnett, G. V., Lane, M.
D., and Kuhajda, F. P., Science 288, 2379–2381 (2000).]
38. (a) In the absence of the phosphatase, the intracellular domain of the insulin receptor remains phosphorylated longer.
The IRS-1 substrate may also retain its phosphate group,
enabling it to stimulate the intracellular processes that allow
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glucose to be brought into cells via the GLUT4 receptor. In
the absence of phosphatase activity to turn off the signaling
pathway, the action of insulin is potentiated, and a normal
physiological effect is observed with a lower concentration
of the hormone.
(b) Injecting the mice with insulin would produce greater
phosphorylation of the insulin receptor and possibly IRS-1,
since there is no phosphatase enzyme to remove the phosphate groups.
(c) Compounds that inhibit the activity of the PTP-1B
phosphatase might be effective in treating diabetes. A concern would be the specificity of these compounds, since unpredictable side effects would result if the drug inhibited
other cellular phosphatases or if PTP-1B has other cellular
targets unrelated to insulin signaling.
40. The bypass surgery eliminates a portion of the stomach and
the small intestine, which are normally sources of hormones that
regulate fuel metabolism and insulin sensitivity. The surgery
apparently alters hormonal signaling enough to restore insulin
sensitivity.
42. Metformin activates AMPK and, in so doing, promotes the
phosphorylation of acetyl-CoA carboxylase, the enzyme that
catalyzes the first committed step of lipid synthesis (see Section
17-2). Phosphorylation inactivates acetyl-CoA carboxylase, which
decreases the concentration of malonyl-CoA and inhibits lipid
synthesis. The decrease in malonyl-CoA relieves the inhibition of
fatty acid transport and allows b oxidation to occur. The result is
a decrease in the concentration of the plasma triacylglycerols that
predispose the patient to atherosclerosis. Phosphorylation of protein kinase B by AMPK would increase translocation of GLUT4
transporters to the plasma membrane. In this manner, AMPK is
acting as insulin does, promoting the uptake of glucose into muscle
and adipose tissue. Stimulation of AMPK by metformin benefits
the patient with metabolic syndrome, who is resistant to insulin,
because AMPK triggers the same responses as insulin signaling.
[From Hardie, D. G., FEBS Lett. 582, 81–89 (2008).]
Chapter 20 Solutions
2. (a) In all the variants, a neutral or negatively charged amino
acid has been replaced with a positively charged amino acid
(Lys or Arg). The 11 abbreviation means that one additional
positive charge was introduced, a 12 indicates the introduction of two additional positive charges, and so on. DNA is
negatively charged because of its phosphodiester backbone.
It’s reasonable to hypothesize that an enzyme with an increased number of positive charges might bind more effectively to the negatively charged DNA.
(b) Intact, double-stranded DNA has a lower absorbance at
260 nm than does single-stranded DNA. An increase in absorbance at 260 nm over time is a useful measurement of the
catalytic activity of DNase, since the products of the reaction
are short, single-stranded oligonucleotides.
(c) All of the variants have lower KM values than the wildtype DNase, indicating that the DNase variants bind more
tightly to the DNA substrate than does the wild-type enzyme.
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The tighter binding is no doubt due to the additional positive charges on the variants, which allow the formation of ion
pairs between the variant enzymes and the DNA. There seems
to be a rough correlation between the number of positively
charged residues and the K M value: An increase in the number of positively charged residues results in a lower K M value,
which indicates tighter binding. However, the Vmax values
must also be assessed. All of the variants have a higher Vmax
value than the wild-type enzyme, indicating that the tighter
binding leads to greater catalytic activity. The replacement of
three amino acids yields a variant enzyme in which the K M
and Vmax values are optimized. When four or five amino acids
are replaced, the velocity is lower than when only three amino
acids are replaced. Since the KM value for the 14 and 15 variants is also generally smaller, it is possible that the enzyme and
substrate bind to each other so tightly that catalytic efficiency
is compromised.
(d) The plasmid DNA normally exists in a supercoiled
circle, as shown in the control lane. The wild-type DNase
can nick the DNA on one strand to convert the plasmid
to the relaxed circular DNA. The 13 mutant is the best of
the three in its ability to cleave supercoiled DNA. All three
mutants are able to produce linear DNA, whereas the wildtype DNase does not. This indicates that the variants can cut
both strands, whereas the wild-type enzyme cuts only one
strand.
(e) All of the mutants have a greater nicking activity than the
wild-type for low- and high-molecular-weight DNA at low
concentrations. The 12 mutant is especially active under
these conditions and would be a good choice to use for a
lupus patient. The 11 and 12 mutants can degrade lowmolecular-weight DNA at a high DNA concentration better
than the wild-type enzyme, but the 13 and 14 mutants
are not as active. Not all of the data are given, so it is difficult to make conclusions about the ability of the mutants
to degrade high-molecular-weight DNA, but the 12 mutant is clearly more effective than the wild-type DNase and
would be a good choice to use for a CF patient. [From Pan,
C.Q., and Lazarus, R.A., J. Biol. Chem. 273, 11701–11708
(1998).]
4. The origin is more likely to be richer in A:T base pairs, since
these experience fewer stacking interactions and are more easily
separated, which would allow easier access for the replication
proteins.
6. Helicase would unwind the parental DNA, exposing extensive segments of single-stranded DNA. Without a polymerase to
convert these template strands into double-stranded DNA, the
single strands would be susceptible to endonucleases, especially
if the supply of replication protein A was limited.
8. The drug would inhibit DNA synthesis because the polymerization reaction is accompanied by the release and hydrolysis
of inorganic pyrophosphate (PPi ; see Fig. 20-9). Failure to hydrolyze the PPi would remove the thermodynamic driving force
for the overall process.
10. DNA polymerase ´ would need to have greater processivity because it synthesizes the leading strand continuously.
48
| CHAPTER 20 Solutions
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Polymerase a can be less processive because it synthesizes only a
short DNA segment at the start of each Okazaki fragment.
12. (a) The positively charged Mg21 ion could decrease the affinity of the DNA’s 39 O atom for H, thereby increasing the
nucleophilicity of the 39 O atom (making it behave more
like an O2 ion). The Mg21 could also help neutralize the
negative charge on the incoming nucleotide.
(b) The Mg21 could stabilize the negative charge that develops on the pentacovalent transition state.
(DNA)
C3
O
O
O
P
P
O
O
O
O
O
O
P
O
C5 (NTP)
O
14. The enzyme most likely detects the geometry of the polynucleotide chain as it shifts from the wide and shallow A-form
characteristic of an RNA helix to the narrower and steeper
B-form of DNA.
16. The DNase can create a nick (a break in one strand). Then
the exonuclease activity of DNA polymerase I can remove nucleotides on the 59 side of the nick. At the same time, the polymerase active site can add radioactive deoxynucleotides to the
39 side of the nick. The removal and replacement of nucleotides
translates the nick in the 59 S 39 direction. DNA ligase can then
seal the gap between the original DNA and the newly synthesized radioactive segment.
18. Eukaryotic DNA ligases use ATP, and bacterial ligases use
NAD1, as a cofactor in the reaction. A drug that inhibited only
NAD1-dependent ligases would be an effective drug to treat
bacterial diseases without harming the host. Bacteria unable to
carry out the ligase reaction cannot complete DNA replication
and cannot survive.
20. Because there is a strong link between telomerase activity
and aging, it’s possible that cancer cells might overexpress the
telomerase enzyme. Increased telomerase activity could contribute to the growth of cancer cells. If the G quartet acts as a negative regulator of telomerase activity, inducing its formation could
inhibit the telomerase and inhibit growth of cancer cells.
22. Both of these proteins bind single strands of DNA: Replication
protein A functions as an SSB protein during DNA replication, and
POT1 binds the single-stranded DNA extension of a telomere.
NH2
24.
H
O
N
N
O
N
N
H
N
N
N
N
H
N
H
Adenine
8-Oxoguanine
26. The structures of the intercalating agents resemble A:T
and G:C base pairs, which explains why they are able to slip in
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between the stacked base pairs of DNA. This has the effect of
creating what appears to the replication machinery as an “extra”
base pair. An extra base incorporated into the newly synthesized
DNA may eventually lead to a frameshift mutation (in which
the additional nucleotide causes the translation apparatus to read
a different set of successive three-nucleotide codons).
28. The ultraviolet light is an additional precaution against contamination. Ultraviolet light causes the formation of thymine
dimers in bacterial DNA. High levels of exposure to ultraviolet
light would overwhelm the bacteria’s ability to repair the dimers,
which would result in the eventual death of the bacteria. Thus,
the ultraviolet light helps keep the hood space free from bacteria
that could contaminate the cultured cells.
30. (a)
O
N
NH
N
O
N
H
Xanthine
(b) Since cytosine also pairs with guanine, the deamination
of guanine to xanthine does not induce a mutation.
32. (a)
NH2
H3C
O
H3C
N
NH
O
N
H
N
H
5-Methylcytosine
O
Thymine
(b) Thymine
(c) A C:G to T:A transition occurs.
(d) The cell cannot repair the deaminated 5-methylcytosine
since the resulting base is indistinguishable from thymine
that occurs normally.
34. (a)
O
N
N
CH3
O
H N
N
N
CH3
N
H
N
O
H
O 6-Methylguanine
Thymine
The methylation of guanine causes a G:C to A:T transition.
H
(b)
H N
H3C
O
N
N
H
N
N
N
H
H
O 6-Methylguanine
36. The thymine dimers form when DNA is exposed to ultraviolet light, so the enzymes that repair the damage are activated
by the same stimulus that produced the damage.
38. The amino acid code is degenerate so that several codons
may specify the same amino acid. For example, the codons GCA,
GCC, GCG, and GCU all code for alanine. So if a mutation occurred at the third position (the 39 end) of the codon, alanine
would still be incorporated into the protein.
40. (a) The polymerases that lack 39 S 59 exonuclease activity
have higher error rates than polymerases that contain the
exonuclease activity.
(b) a, once every 6250 bases; b, once every 1493 bases; d,
once every 100,000 bases; ´, once every 100,000 bases; h,
once every 29 bases.
(c) Polymerase h and HIV RT incorporate the correct base
with approximately the same efficiency (for example, 420,
760, and 800 mM ? min21 3 103). However, polymerase
h incorporates mispaired bases much more efficiently than
HIV RT does (22, 1.6, and 8.7 vs. 0.07 mM ? min21 3 103).
These results indicate that the high error rate of polymerase
h results from its ability to incorporate the wrong base rather
than its inability to incorporate the correct base.
(d) Overexpression of an error-prone DNA polymerase
similar to polymerase h would increase the mutation rate.
[From Matsuda, T., Bebenek, K., Masutani, C,. Hanaoka,
F., and Kunkel, T.A., Nature 404, 1011–1013 (2000).]
42. A solution of 0.5 M NaCl effectively dissociates the positively
charged histone proteins from the negatively charged DNA by
disrupting the ionic bonds that hold these two chromatin components together. The charged sodium and chloride ions can form
substitute ionic interactions with the histone protein and DNA.
44. The high degree of sequence conservation from cows to peas
indicates that the sequence of the histone H4 protein is so vital
to its function that amino acid substitutions, especially those
that are nonconservative in nature, would disrupt the function
of the protein and thus cannot be tolerated.
46. DNA methylation is important in the process of imprinting, in which DNA expression depends on parental origin. In
the absence of DNA methylation, imprinting cannot take place.
DNA methylation is important in gene expression during embryonic development, and if the DNA cannot be methylated,
the embryos do not develop properly and die before birth.
N
O
N
No, since G normally base pairs with C, a mutation does
not result. A mutation occurs only when O 6-methylguanine
pairs with T. [From Leonard, G.A., Thomson, J., Watson,
W.P, and Brown, T., Proc. Natl. Acad. Sci. 87, 9573–9576
(1990).]
Protonated cytosine
Chapter 21 Solutions
2. Organizing genes with related functions in an operon means
that the genes can be turned on and off together. Furthermore,
because the genes are transcribed as a unit, the products can be
generated in roughly equivalent concentrations and in the same
region of the cell (where the ribosome is located). Consequently,
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only one signal is needed to activate or deactivate the metabolic
function carried out by the products of the operon’s genes. A
eukaryotic cell can coordinate the transcription of physically
separated but functionally related genes if they are share a set of
enhancers, silencers, activators, or repressors.
4. The protein must identify a gene to be expressed and must
activate the transcription machinery so that the DNA can be
transcribed to RNA. The gene is identified by the ability of the
protein to bind specific DNA sequences associated with the gene.
The resulting distortion in the DNA can then be recognized in
a sequence-independent manner by transcription factors that assemble at the site in order to initiate RNA synthesis.
6. RNA polymerase I transcribes rRNA genes. Although there
are multiple copies of these genes, there are only a few types of
rRNA molecules and the cell requires them in equal amounts.
Therefore, rRNA gene transcription does not need to be selective, and one promoter suffices. In contrast, the protein-coding
genes transcribed by RNA polymerase II must be expressed at
different times and at different levels. The regulation of transcription of these genes is more elaborate, requiring different
promoter sequences that can potentially interact with different
sets of transcription factors.
8.
11
GAGCATATAAGGTGAGGTAGGATCAGTTGCTCCTCACATTT
TATA box
10. The other product is methylamine.
Inr
O
H
N
CH
O
H
N
C
CH2
CH2
NH
C
C
CH2
H2O
CH2
CH2
Arginine
CH
H2N
CH3
Methylamine
CH2
Citrulline
N CH3
H
NH2
NH
C
O
NH2
12. The presence of the s factor decreases the affinity of RNA
polymerase for DNA. This allows the polymerase–s factor complex to quickly scan long segments of DNA for promoter sequences. Once transcription has begun, the s factor is no longer
needed and dissociates from the enzyme. Now the RNA polymerase has a high affinity for DNA, which helps keep it associated with the template during transcription.
14. (a) When glucose is present, cAMP concentration is low.
CAP cannot bind to the operon in the absence of cAMP,
so the operon is off. Glucose is the preferred substrate, so
lactose is not used as long as glucose is present.
(b) The operon is off, both because glucose is present and
CAP is not bound [see part (a)] and because lactose is absent and the repressor is bound to the operator, preventing
RNA polymerase from binding.
(c) CAP binds cAMP and can bind to the operon, but the
operon remains off because the repressor is also bound in
the absence of lactose.
50
| CHAPTER 21 Solutions
(d) The operon is on. In the absence of glucose, cAMP levels
rise. cAMP binds to CAP, which can bind to the operon to
stimulate transcription. In the presence of lactose, allolactose
is formed and binds to the repressor, removing the repressor
from the operator. The genes of the operon are expressed and
the bacterial cell can use lactose as a food source.
16. If the repressor can bind to the operator, the genes of the
operon are not expressed. When lactose is added to the growth
medium, it cannot bind to the repressor. The repressor remains
bound to the DNA and the genes of the operon are not expressed. This is an example of a noninducible mutation, because
adding lactose does not result in gene expression.
18. When tryptophan is plentiful, there is “extra” tryptophan
available to bind to the repressor. With tryptophan bound, the
repressor binds to the promoter and prevents RNA polymerase
from binding. The genes of the trp operon are not expressed
because tryptophan biosynthetic enzymes are not needed when
tryptophan is plentiful. But when the concentration of tryptophan drops, tryptophan dissociates from the repressor, causing
a conformational change that results in the dissociation of the
repressor from the promoter. RNA polymerase can now bind to
the promoter, and the genes encoding the enzymes for the tryptophan synthetic pathway can now be expressed.
20. The 59 end of any prokaryotic RNAs ending in A will be
labeled. The 59 ends of RNA transcripts have 59 triphosphate
groups containing the labeled g-phosphate. The phosphodiester
bonds in the RNA will not be labeled because these phosphate
groups come from the a-phosphates of the nucleoside triphosphates (the g-phosphates are released as pyrophosphate).
*pppA 1 pppN z
y *pppApN 1 PPi
22. The incorporation of cordycepin into a growing RNA
chain provides evidence that transcription occurs in the 59 S 39
direction. Once incorporated, transcription halts because of the
lack of a 39 OH group. If transcription occurred in the 39 S 59
direction, cordycepin would not be incorporated into the growing
RNA chain and would not be able to halt RNA synthesis.
24. The prokaryotic RNA polymerase is a five-subunit enzyme
with a subunit composition of a2bb9vs and is the only RNA
polymerase in prokaryotic cells. In contrast, eukaryotic cells have
three different RNA polymerase enzymes that differ in structure from the prokaryotic polymerase. Thus, rifampicin can
be effectively used as an antibiotic to treat bacterial diseases in
eukaryotic organisms because the drug is able to block transcription in the disease-causing prokaryotic cells without inhibiting
transcription in the eukaryotic organism.
26. (a) mRNA(n residues) 1 Pi S NDP 1 mRNA(n 2 1 residues)
(b) The reverse of the phosphorolysis reaction is an RNA
polymerization reaction. PNPase uses an NDP substrate to
extend the RNA by one nucleotide residue and releases Pi.
RNA polymerase uses an NTP substrate and releases PPi.
(c) High processivity would allow the exonuclease to rapidly degrade mRNA molecules. This would be important
in cases where the gene product was no longer needed. An
mRNA that was degraded more slowly could potentially
continue to be translated.
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28. The C-terminal domain (CTD) is phosphorylated when RNA
polymerase transitions from initiation mode to elongation mode. If
the CTD is missing, it cannot be phosphorylated, and elongation
cannot occur. Transcripts that do manage to clear the promoter in
the absence of a phosphorylated domain will not be properly processed, since the phosphorylated CTD serves as a docking site for
enzymes involved in cotranscriptional modification.
30. (a) 59…NNAAICICCINNNNCCIICICUUUUUUNNN…39
(b)
N
N
N
N
5
I
C
C
I
C
I
A
A
NN
C
C
I
I
C
I
C
U
U
UUUU
3
The RNA terminator hairpin is much less stable when I is
substituted for G because I:C base pairs have only two hydrogen bonds whereas G:C base pairs have three. The stability of
the RNA hairpin is important in termination. When ITP is
substituted for GTP in culture, the hairpin might not form; as
a consequence, RNA transcripts will not terminate properly.
32. The b,g-imido nucleoside triphosphate can be used as a
substrate by RNA polymerase, since it is the phosphoanhydride
bond closest to the ribose that is cleaved when the nucleotide
is incorporated into the growing RNA transcript. Elongation is
therefore not affected by the presence of the modified nucleotide.
Rho-dependent termination is affected because Rho is a helicase
that uses the energy of ATP hydrolysis to pry apart the DNA–
RNA hybrid. The b,g-imido nucleoside triphosphate cannot
be hydrolyzed to its corresponding diphosphate and inorganic
phosphate, so Rho-dependent termination cannot occur.
34. (a) CAG codes for Gln. The resulting protein would contain a series of extra Gln residues. These polar residues
would most likely be located on the protein surface but
could interfere with protein folding, stability, interactions
with other proteins, and catalytic activity.
(b) The longer transcripts could be due to transcription initiating upstream of the normal site or failing to stop at the
usual termination point. Longer mRNA molecules could
also result from the addition of an abnormally long poly(A)
tail or the failure to undergo splicing.
(c)
G C
A
A
C G
G C
A A
C G
G C
A A
C G
CAG CAG
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[From Fabre, E., Dujon, B., and Richard, G.-F., Nuc. Acids
Res. 30, 3540–3547 (2002).]
36. The capped mRNA has a 59–59 triphosphate linkage, which
is not recognized by exonucleases (which normally cleave 59–39
phosphodiester bonds). Capping must occur as soon as the 59
end of the mRNA emerges from the RNA polymerase so that the
message will not be degraded by exonucleases.
38. (a) The active site of poly(A) polymerase is narrower because it does not need to accommodate a template strand.
(b) Poly(A) polymerase binds only ATP, whereas other RNA
polymerases bind ATP, CTP, GTP, and UTP.
40. Bacterial genes contain no introns, so the cells, which lack
splicing machinery, cannot express eukaryotic genes containing introns and exons. Mature eukaryotic mRNA, which has already been
spliced, contains only exons, which, after they have been converted
back to DNA, can be transcribed and translated by the bacteria.
42. The mRNA from a gene may be alternatively spliced to
yield several different types of proteins. This increases the diversity of the proteins produced by the cell without a correspondingly large number of genes.
44.
NH
S
2
HN
O
N
S
N
N
Ribose
Ribose
4-Thiouridine
2-Thiocytidine
46. The primary structure of a protein refers to its sequence of
amino acids; in the RNase P RNA this corresponds to the sequence of 417 nucleotides. Secondary structure in proteins refers
to regular, repeating structural motifs such as a helices and b
sheets. In the ribozyme, secondary structure refers to the basepaired stem and loop structures. Tertiary structure in proteins
refers to the overall three-dimensional shape of the macromolecule; similarly for the ribozyme, the tertiary structure refers to
the three-dimensional shape of the molecule.
Chapter 22 Solutions
2. Because the genetic code is degenerate, a mutation that alters a
codon may not alter the amino acid encoded by that codon. This
is particularly true for mutations at the third codon position.
Changes at the first or second position almost always change the
encoded amino acid, but the new amino acid may be chemically
similar to the old one (for example, Val S Ala).
4.
H
O
N
N
N
H
H
N
N
N
N
N
N
I:A
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6. (a) The RNA-binding site of the synthetase is necessary for
binding tRNA for aminoacylation and for binding RNA
during splicing.
(b) Because of their long evolutionary history, aminoacyl–
tRNA synthetases have had many opportunities to diversify
in structure and in function in order to assume additional
roles in the cell.
8. (a) The enzyme itself must act as a template to direct the addition of two C residues followed by one A residue.
(b) The enzyme must recognize only ATP and CTP as substrates, excluding GTP, UTP, and all dNTPs.
(c) In the two-domain enzyme, one polymerase adds the two
C residues, and the other domain then adds the terminal A
residue.
10. In order for a cell to incorporate a nonstandard amino acid
into a polypeptide, the amino acid must first be attached to a
tRNA corresponding to one of the 20 standard amino acids.
The aminoacyl–tRNA can then bind to the ribosome and its
amino acid can be incorporated into the growing polypeptide
at positions corresponding to the codon for the standard amino
acid. The failure of cells to synthesize norleucine-containing
peptides most likely reflects the inability of LeuRS to efficiently attach norleucine to tRNALeu. A mutant LeuRS, which
presumably lacks the proofreading activity of the wild-type
LeuRS, was able to produce norleucine–tRNALeu, and the
cells’ ribosomes used this aminoacylated tRNA to translate
Leu codons.
12. The assembly of functional ribosomes requires equal amounts
of the rRNA molecules. Therefore, it is advantageous for the cell
to synthesize the rRNAs all at once.
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in the A site. Changing one of these two rRNA residues would
inactivate the translational proofreading mechanism by eliminating the specific hydrogen bonding between the rRNA and
the mRNA. As a result, incorrectly paired tRNAs could not be
distinguished from correctly paired tRNAs, and the error rate of
translation would increase.
22. The ribosome minimizes the chances of misreading the
A-site codon by binding the A-site tRNA with lower affinity. If
the tRNA bound with higher affinity, it would be less likely to
dissociate as part of the proofreading mechanism.
24. If EF-Tu formed a complex with fMet–tRNA fMet, the
fMet–tRNA could be delivered to the ribosomal A site when
a Met codon was positioned there. However, transpeptidation
could not occur because the amino group of fMet is blocked
by the formyl group. Polypeptide synthesis would be halted
until the fMet–tRNA fMet was replaced by Met–tRNAMet in the
A site.
26. (a) Translation begins at the first AUG codon (ATG in the
DNA). The polypeptide sequence is Met–Val–His–Leu–
Thr.
(b) The mutated sequence has a T residue inserted in the
second codon. This is a frameshift mutation, so all codons
following that point will be altered. The polypeptide sequence is Met–Val–Ala–Ser–Asp.
28. The number of phosphoanhydride bonds (about 30 kJ ?
mol21 each) that are cleaved in order to synthesize a 20-residue polypeptide can be calculated as follows (the relevant
ATP- or GTP-hydrolyzing proteins are indicated in parentheses):
Aminoacylation (AARS)
2 3 20 ATP
14. Ribosomal inactivating proteins catalyze the removal of adenine residues from ribosomal RNA. This is analogous to the
removal of a side chain from an amino acid residue in a protein.
Like proteins, ribosomal RNAs have specific residues that are
essential to their function; removing these residues causes loss
of activity.
Translation initiation (IF-2)
1 GTP
Positioning of each aminoacyl–tRNA (EF-Tu)
19 GTP
16. The peptidyl transferase activity lies entirely within the
23S rRNA; that is, 23rRNA is a ribozyme. The proteins might
be necessary to assist the 23S rRNA in forming the necessary
three-dimensional structure required for catalytic activity, just
as the proper conformation is required for protein enzymes.
Extremely strong intermolecular interactions between the
proteins and the rRNA confirm the importance of the proteins and explain why the extraction process failed to remove
them.
Thus, approximately 80 3 30 kJ ? mol21, or 2400 kJ, is required.
In a cell, proofreading during aminoacylation and during translation requires the hydrolysis of additional phosphoanhydride
bonds, making the cost of accurately synthesizing the 20-residue
polypeptide greater than 2400 kJ ? mol21.
30. If a peptidyl–tRNA dissociates from the ribosome during
translation, the hydrolase releases the peptide from the tRNA.
Because peptide synthesis is prematurely terminated, the polypeptide is likely to be nonfunctional, and its amino acids must
be recycled. Similarly, the tRNA, once released from the peptidyl
group, can be reused. The essential nature of the peptidyl–tRNA
hydrolase suggests that ribosomes that have initiated translation
sometimes stop translating before reaching a stop codon.
32. (a) Transpeptidation involves the nucleophilic attack of the
amino group of the aminoacyl–tRNA on the carbonyl carbon of the peptidyl–tRNA (see Fig. 22-15). The higher the
pH, the more nucleophilic the amino group (the less likely
it is to be protonated).
18. S1 helps to maintain mRNA in the single-stranded state
and prevents it from forming a double-stranded structure that
would block initiation of translation because the initiator tRNA
would be unable to bind.
20. In 16S rRNA, A1492 and A1493 act as a sensor to distinguish correctly and incorrectly paired codons and anticodons.
tRNA binding triggers a conformational change in the rRNA
that allows A1492 and A1493 to form hydrogen bonds with an
mRNA that has correctly base paired with a tRNA anticodon
52
| CHAPTER 22 Solutions
Translocation after each transpeptidation (EF-G)
19 GTP
Termination (RF-3)
1 GTP
Total: 80 ATP equivalents
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(b)
NH
Peptidyl–tRNA
CHR
O
H
C
O
tRNA
N H
CHR
Aminoacyl–tRNA
O
C
O
tRNA
(c)
NH2
N
HN
N
N
The protonated A2451 residue, which has a positive charge,
could stabilize the negatively charged oxyanion of the tetrahedral reaction intermediate. As the pH increases, A2451
would be less likely to be protonated, so this catalytic mechanism would be less effective at higher pH.
34. (a) Introduction of a stop codon would terminate protein
synthesis prematurely, so no functional b-galactosidase
would be synthesized. If the stop codon were located near
the C-terminus of the protein, the polypeptide would be
shorter than normal but might still retain activity if its active site region were intact.
(b) These results indicate that the oxazolidinone increases
the ability of ribosomes to overlook the stop codon and synthesize b-galactosidase polypeptides of normal size.
(c) b-Galactosidase activity would be extremely low because a
nucleotide insertion or deletion changes the reading frame for
translation. The resulting polypeptide would have a different
amino acid sequence and would therefore be nonfunctional.
(d) These results indicate that the oxazolidinone promotes
frameshifting in the ribosome so that despite the insertion
or deletion, the correct reading frame is occasionally translated and some functional b-galactosidase is synthesized.
(e) There are two possible Glu codons: GAA and GAG.
Changing a single base in these codons generates codons
specifying four different amino acids. Substitutions at the
second position yield codons for Ala (GCA and GCG), Gly
(GGA and GGG), and Val (GUA and GUG). Substitution
at the first position yields codons for Gln (CAA and CAG).
(f ) The oxazolidinone does not promote codon misreading,
that is, incorporation of an amino acid other than the one
specified by a codon. If codon misreading were occurring, the
encoded Ala, Gln, Gly, and Val codons would occasionally be
read as Glu codons, and a functional enzyme would result.
(g) The ability of the oxazolidinone to cause the ribosome
to read through stop codons would result in the synthesis
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of longer-than-normal polypeptides. These proteins might
not fold properly or might not be catalytically active, which
would interfere with normal cellular metabolism. The ability
of the oxazolidinone to promote frameshifting would result
in polypeptides with garbled amino acid sequences, which
would be nonfunctional and potentially toxic to the cell.
(h) No; the oxazolidinone affects translational accuracy,
which is primarily a function of the 30S ribosomal subunit, the location of the binding sites for mRNA and the
tRNA anticodon loop. A binding site near the peptidyl
transferase site would be more consistent with an effect
on peptide bond formation. Presumably, communication
between the 50S and 30S subunits allows binding at one
site to influence events at another site. [From Thompson,
J., O’Connor, M., Mills, J.A., and Dahlberg, A.E., J. Mol.
Biol. 322, 273–279 (2002).
36. Protein folding is driven by the hydrophobic effect (see Section 4-3), which directs nonpolar side chains to the interior of
the folded protein because these side chains do not interact favorably with water. The derivatized protein was able to fold as
easily as the native protein because the lysine residues are located
on the surface of the protein and do not play a large role in the
protein-folding process.
38. Cells expressing growth factor receptors that have lost the
ligand-binding domain but retain the tyrosine kinase domain are
transformed cells; that is, they are cancerous cells (see Box 10-B)
because they grow and proliferate in the absence of extracellular
signaling ligands. Hsp90 binds to the tyrosine kinase domain to
stabilize it or help it fold. When geldanamycin is added to these
transformed cells, Hsp90 can no longer function, the tyrosine
kinase activity drops, and the cells no longer exhibit the characteristics of transformed cells.
40. (a) The a chains, when in excess, combine with all available b
chains to form functional a2b2 hemoglobin, thereby minimizing the formation of nonfunctional b4 hemoglobin.
(b) The protein helps prevent the precipitation of the a
chains that have not yet paired with b chains.
(c) When a deficiency of b chains is coupled with an excess
of a chains, the a chains precipitate and destroy the red
blood cells, worsening the anemia that results from the lack
of b chains.
(d) The imbalance between the amounts of a and b chains is
minimized when the synthesis of both globins is depressed due
to mutations in both an a globin gene and a b globin gene.
(e) The process of initiating translation requires that eIF2
hydrolyze its bound GTP to GDP and Pi . In order for
the protein to participate in subsequent translation initiation events, its GDP must be replaced with GTP. If this
exchange does not occur, reinitiation is not possible, and
protein synthesis comes to a halt.
(f ) Heme prevents the phosphorylation of eIF2, so translation initiation can proceed. This mechanism regulates
the level of globin synthesis according to the availability of
heme. Consequently, the cells can produce functional hemoglobin, which contains globin polypeptides as well as
heme prosthetic groups.
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42. During translation, the SRP must be positioned on the ribosome where the signal peptide emerges from the exit tunnel. The
SRP then undergoes a conformational change that temporarily
pauses the translation process. In order to fulfill these roles, the
SRP must be able to bind to the ribosome, most likely through
interactions between the SRP RNA and ribosomal RNA.
44. If the SRP waited to bind the polypeptide until translation
was complete, it is highly likely that the newly synthesized protein
would have already misfolded into a conformation that included
it signal peptide or would have aggregated with other cellular proteins. By binding to the nascent polypeptide as soon as its signal
sequence emerges from the ribosome, the SRP ensures the delivery of the ribosome to the ER membrane so that the polypeptide
can be translocated co-translationally. Chaperones within the ER
lumen then ensure that the protein folds properly.
46. The SRP recognizes the N-terminal signal sequence as it
emerges from the ribosome, pauses translation, and escorts the
entire complex to the ER membrane, where it docks with its
receptor and translation resumes. GTP is required for this process. GTP hydrolysis could be part of a proofreading step by
occurring in response to SRP conformational changes and by
triggering additional changes in the SRP. This mechanism would
ensure that only proteins with signal sequences were translocated
into the ER. Cytosolic proteins lacking signal sequences would
not trigger GTP hydrolysis, and the SRP would not be able to
dock with its ER receptor.
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48.
HO
H
CH2OH
O
H
OH H
H
H
O
CH3 NH
CH
C
NH
C
CH
O
O
CH3
50. G protein–coupled receptors (see Section 10-2) have palmitoylated Cys residues and as such are classified as lipid-linked
proteins. The nonpolar palmitate acyl chain helps to anchor this
integral membrane protein to the membrane.
O
H
N
CH
C
OH
CH2
S
O
C
(CH2)14
CH3