Furthermore, covalent modification allows cells to escape allosteric regulation by covalently locking the modified enzyme in an active or inac-tive state, regardless of effector concentrat
Trang 1In (a), Vmaxdoubles, but K mis constant.
In (b), Vmaxhalves, but K mis constant
In (c), Vmaxis constant, but the apparent K mincreases
In (d), Vmaxdecreases, but K mis constant
In (e), Vmaxdecreases, K m decreases, but the ratio K m /Vmaxis
constant
6 a The slope is given by K m /Vmax(KSA/[A] 1)
b y-intercept ((K m /[A]) 1)1/Vmax
c The horizontal and vertical coordinates of the point of
intersection are 1/[B] K m /KSAK m and
1/v 1/Vmax(1 (K m /KSA))
7 Top left: (1) Competitive inhibition (I competes with S for
binding to E) (2) I binds to and forms a complex with S
Top right: (1) Pure noncompetitive inhibition (2) Random,
single-displacement bisubstrate reaction, where A doesn’t affect
B binding, and vice versa (Other possibilities include [3]
Irre-versible inhibition of E by I; [4] 1/v vs 1/[S] plot at two
differ-ent concdiffer-entrations of enzyme, E.)
Bottom left: (1) Mixed noncompetitive inhibition (2) Ordered
single-displacement bisubstrate mechanism
Bottom right: (1) Uncompetitive inhibition (2)
Double-displacement (ping-pong) bisubstrate mechanism
8 Clancy must drink 694 mL of wine, or about one 750-mL bottle
9 a KS m cat 5 103sec1; d kcat/K m
1.67 108M1sec1; e Yes, because kcat/K mapproaches the
limiting value of 109M1sec1; f Vmax 105mol/mL sec;
g [S] maxwould equal 2 105mol/mL sec, but
K m
10 a Vmax 1sec1; b kcat 44,000 sec1; c kcat/K m
24.4 108M1sec1; d Yes! kcat/K mactually exceeds the
theoreti-cal limit of 109M1sec1in this problem; e The rate at which E
encounters S; the ultimate limit is the rate of diffusion of S
11 a Vmax 1sec1; b v 1sec1;
c kcat/K m 1.6 108M1sec1; d Yes
(e)
v
[S]
O
I
(d)
v
[S]
O
I
(c)
v
[S]
O
I
12 a Vmax 1sec1; b kcat 1.2 106sec1;
c kcat/K m 1 108M1sec1; d Yes! kcat/K mapproaches the theoretical limit of 109M1sec1; e The rate at which E en-counters S; the ultimate limit is the rate of diffusion of S
13 a Vmax 1sec1; b kcat 1.28 104sec1;
c kcat/K m 1.4 108M1sec1; d Yes! kcat/K mapproaches the theoretical limit of 109M1sec1
14 a Vmax 120 mmol mL1sec1; b v 104.7 mmol mL1sec1;
c kcat/K m 3.64 108M1sec1; d Yes! kcat/K mapproaches the theoretical limit of 109M1sec1
15 a Starting from Vmaxf k2[ET ] and Vmaxr k1[ET] for the maximal rates of the forward and reverse reactions
respec-tively, and the Michaelis constants K mS(k1 k2)/k1and
K mP(k1 k2)/k2for S and P, respectively,
v (Vmaxf[S]/K mS)Vmaxr[P]/K mp)/(1 [S]/K mS [P]/K mP)
b At equilibrium, v 0 and Keq [P]/[S] Vmaxf
K mP/VmaxrK mS
16 a S is the preferred substrate; its K m is smaller than the K mfor
T, so a lower [S] will give v Vmax/2, compared with [T]
b kcat/K m defines catalytic efficiency kcat/K mfor S
2 107M1sec1; kcat/K mfor T 4 107M1sec1, so the enzyme is a more efficient catalysis with T as substrate
17 a Because the enzyme shows maximal activity at or below 40°C,
it seems more like a mammalian enzyme than a plant enzyme, which would be expected to have a broader temper-ature optimum because plants experience a broader range of temperatures
b An enzyme from a thermophilic bacterium growing at 80°C would show an activity versus temperature profile similar to this one but shifted much farther to the right
Chapter 14
1 a Nucleophilic attack by an imidazole nitrogen of His57on the OCH2O carbon of the chloromethyl group of TPCK
cova-lently inactivates chymotrypsin (See The Student Solutions Manual, Study Guide and Problems Book for structures.)
b TPCK is specific for chymotrypsin because the phenyl ring of the phenylalanine residue interacts effectively with the binding pocket of the chymotrypsin active site This positions the chloromethyl group to react with His57
c Replacement of the phenylalanine residue of TPCK with arginine or lysine produces reagents that are specific for trypsin
2 a The structures proposed by Craik et al., 1987 (Science
237:905–907) are shown here (If you look up this reference, note that the letters A and B of the figure legend for Figure
3 of this article actually refer to parts B and A, respectively Reverse either the letters in the figure or the letters in the figure legend and it will make sense.)
O
N
H
H
H H O
O
O
N H
G Ser 195
56 N
Asp102
Ser214
His57
A
Trang 2
b Asn102of the mutant enzyme can serve only as a
hydrogen-bond donor to His57 It is unable to act as a hydrogen-bond
acceptor, as aspartate does in native trypsin As a result, His57
is unable to act as a general base in transferring a proton
from Ser195 This presumably accounts for the diminished
ac-tivity of the mutant trypsin
3 a The usual explanation for the inhibitory properties of
pep-statin is that the central amino acid, pep-statine, mimics the
tetra-hedral amide hydrate transition state of a good pepsin
substrate with its unique hydroxyl group
b Pepsin and other aspartic proteases prefer to cleave peptide
chains between a pair of hydrophobic residues, whereas
HIV-1 protease preferentially cleaves a Tyr-Pro amide bond
Because pepstatin more closely fits the profile of a pepsin
substrate, we would surmise that it is a better inhibitor of
pepsin than of HIV-1 protease In fact, pepstatin is a potent
inhibitor of pepsin (KI 1 nm) but only a moderately good
inhibitor of HIV-1 protease, with a KIof about 1
4 The enzyme-catalyzed rate is given by:
5 This problem is solved best by using the equation derived in
problem 4
e(Gu‡Ge‡)/RT
ke
ku
So
or
Similarly, for the uncatalyzed reaction:
Assuming that ⬵
Then
ke
ku e(G‡G‡ )兾RT
ke
ku e G
‡兾RT
e G ‡兾RT
k e
k u
ku k
ue G‡
u 兾RT
ke k
e e G ‡兾RT
ke[ES] k
ee G‡兾RT[ES]
[EX‡] K ‡[ES] eG‡兾RT[ES]
K ‡ e G‡兾RT
G ‡ RT ln K ‡
K ‡[EX‡] [ES]
v ke[ES] k
e[EX‡]
O
N
N H
H H
H O
O
H
G Ser195
56 N
Asn102
Ser214
His57
B
H
H
Using this equation, we can show that the difference in activa-tion energies for the uncatalyzed and catalyzed hydrolysis reac-tions (Gu Gc) is 92 kJ/mol
6 Trypsin catalyzes the conversion of chymotrypsinogen to
-chymotrypsin, and chymotrypsin itself catalyzes the conversion
of -chymotrypsin to -chymotrypsin.
7 The mechanism suggested by Lipscomb is a general base pathway in which Glu270promotes the attack of water on the carbonyl carbon of the substrate:
8 Using the equation derived in problem 4, it is possible to
calcu-late the ratio ke/kuas 1.86 1012
9 If the concentration of free enzyme is equal to the concentra-tion of enzyme–ligand complex, the concentraconcentra-tion of ligand would be 1 1027M This corresponds to 1.67 103liters per molecule
10.G 154 kJ/mol This value is intermediate between
nonco-valent forces (H bonds are typically 10–30 kJ/mol) and cononco-valent bonds (300–400 kJ/mol)
11 Assuming that the rate of gluconeogenesis would be equal to
the slowest step in the pathway, with k 2 1020/sec, and as-suming a cellular concentration of fructose-1,6-bisphosphatase
of 0.031 (see Table 18.2), the rate of sugar synthesis would be
2 1020/sec 0.031 mM, or 6.2 1022mM/sec Assuming
a total human cell volume of 40 L, this corresponds to 2.48
1023moles glucose/sec Assuming 30 ATP per glucose (under cellular conditions) and 50 kJ/mole of ATP hydrolyzed, we find that the rate of energy production is (2.48 1023) (30 ATP/glucose) (30.5 kJ/mole), or 2.269 1020kJ/sec Converting to kilocalories and years, we find that the time to synthesize the needed 480 kilocalories with an uncatalyzed reac-tion would be 2.8 1015years, or roughly 200,000 times the life-time of the universe so far
12 The correct answer is c The stomach is a very acidic environ-ment, whereas the small intestine is slightly alkaline The lower part of the figure shows that enzyme X has optimal activity near
pH 2, whereas enzyme Y works best at a pH near 8
13 The correct answer is a The two enzymes have nonoverlapping
pH ranges, so it is highly unlikely that they could operate in the same place at the same time
P
O O
O
C
C O O H
N H H
NH BH
COO
O
O CHR
H
R A
O
OC O Gla270
O
A
A A
O O
O
C
C
O
O H NH
HNH COO
O
O CHRO
H
R A A A A
O C
O
OH NH
OCHRO A
Carboxypeptidase
PO O
OC O H Gla270
O
O OC
NH2
COO H
R A A P
Trang 314 The correct answer is b Only enzyme A has a temperature
range that encompasses human body temperature (37°C)
15 The correct answer is d The activities of the two enzymes
overlap between 40° and 50°C
16 The correct answer is c We have no information on the pH
be-havior of enzymes A and B, nor on the bebe-havior of X and Y as a
function of temperature The only answer that is appropriate to
the data shown is c
17 The only possible answer is b, because a “leveling off” implies
that all the enzyme is saturated with S
18 The correct answer is c In order to bring the substrate into the
transition state, an enzyme must enjoy environmental conditions
that favor catalysis There is no activity apparent for enzyme Y
below pH 5.5
Chapter 15
1 a As [P] rises, the rate of P formation shows an apparent
decline, as enzyme-catalyzed conversion of P ⎯→ S becomes
more likely
b Availability of substrates and cofactors
c Changes in [enzyme] due to enzyme synthesis and
degradation
d Covalent modification
e Allosteric regulation
f Specialized controls, such as zymogen activation, isozyme
variability, and modulator protein influences
2 Proteolytic enzymes have the potential to degrade the proteins
of the cell in which they are synthesized Synthesis of these
enzymes as zymogens is a way of delaying expression of their
ac-tivity to the appropriate time and place
3 Monod, Wyman, Changeux allosteric system:
Hanes–Woolf plot
I
O A
[S]
[S]/v
Lineweaver–Burk plot
A
1/[S]
4 Using the curves for negative cooperativity as shown in Figure 15.8 as a guide:
5 For n 2.8, Ylungs 0.98 and Ycapillaries 0.77
For n 1.0, Ylungs 0.79 and Ycapillaries 0.61
Thus, with an n of 2.8 and a P50of 26 torr, hemoglobin becomes almost fully saturated with O2in the lungs and drops to 77%
saturation in resting tissue, a change of 21% If n of 1.0 and a
P50of 26 torr, hemoglobin would become only 79% saturated with O2in the lungs and would drop to 61% in resting tissue, a change of 18% The difference in hemoglobin O2saturation
conditions between the values for n 2.8 and 1.0 (21% 18%,
or 3% saturation) seems small, but note that the potential for
O2delivery (98% saturation versus 79% saturation) is large and
becomes crucial when pO2in actively metabolizing tissue falls below 40 torr
6 More glycogen phosphorylase will be in the glycogen
phos-phorylase a (more active) form, but caffeine promotes the less
active T conformation of glycogen phosphorylase
7 Over time, stored erythrocytes will metabolize 2,3-BPG via the pathway of glycolysis If [BPG] drops, hemoglobin may bind O2
with such great affinity that it will not be released to the tissues (see Figure 15.29) The patient receiving a transfusion of [BPG]-depleted blood may actually suffocate
8 a By definition, when [Pi] K0.5, v 0.5 Vmax
b In the presence of AMP, at [Pi] K0.5, v 0.85 Vmax(from Figure 15.14c)
c In the presence of ATP, at [Pi] K0.5, v 0.12 Vmax(from Figure 15.14b)
9 If G–GTPase activity is inactivated, the interaction between G
and adenylyl cyclase will be persistent, adenylyl cyclase will be active, [cAMP] will rise, and glycogen levels will fall because
[S]
[S]
Negative cooperativity
Negative cooperativity
Negative cooperativity
Michaelis–
Menten
Michaelis–
Menten
Lineweaver–Burk
Hanes–Woolf
M-M
v
1
—
v
1/[S]
[S]
—
v
Trang 4glycogen phosphorylase will be predominantly in the active,
phosphorylated a form.
10 An excess of a negatively cooperative allosteric inhibitor could
never completely shut down the enzyme Because the enzyme
leads to several essential products, inhibition by one product
might starve the cell for the others
11 a
-R(R/K)X(S*/T*)-b
-KRKQIAVRGL-12 Ligand binding is the basis of allosteric regulation, and allosteric
effectors are common metabolites whose concentrations reflect
prevailing cellular conditions Through reversible binding of
such ligands, enzymatic activity can be adjusted to the
momen-tary needs of the cell On the other hand, allosteric regulation is
inevitably determined by the amounts of allosteric effectors at
any moment, which can be disadvantageous Covalent
modifica-tion, like allosteric regulamodifica-tion, is also rapid and reversible,
because the converter enzymes act catalytically Furthermore,
covalent modification allows cells to escape allosteric regulation
by covalently locking the modified enzyme in an active (or
inac-tive) state, regardless of effector concentrations One
disadvan-tage is that covalent modification systems are often elaborate
cascades that require many participants
13 Sickle-cell anemia is the consequence of Hb S polymerization
through hydrophobic contacts between the side chain of Val6
and a pocket in the EF corner of -subunits Potential drugs
might target this interaction directly or indirectly For example,
a drug might compete with Val6 side chain for binding in the
EF corner Alternatively, a useful drug might alter the
conforma-tion of Hb S such that the EF corner was no longer accessible or
accommodating to the Val6 side chain Another possibility
might be to create drugs that deter the polymerization process
in other ways through alterations in the surface properties
between Hb S molecules
14 Nitric oxide is covalently attached to Cys93 Thus, the
interac-tion is not reversible binding, as in allosteric regulainterac-tion On the
other hand, the reaction of NOj with this cysteine residue is
ap-parently spontaneous, and no converter enzyme is needed to
add or remove it Thus, the regulation of covalent modification
that is afforded by converter enzyme involvement is obviated
The Hb⬊NOj interaction illustrates that nature does not always
neatly fit the definitions that we create
15 Pro: Lactate is a metabolic indicator of the need for oxygen; it
binds to Mb at a distinct site (the allosteric site?) and it lowers
Mb’s affinity for O2
Con: Mb is a monomeric protein Traditionally, allosteric
phe-nomena have been considered the realm of oligomeric proteins
How then is Mb “allosteric”? Since a ligand-induced conforma-tional change in a monomeric proteins can affect binding of
“substrate” (i.e., O2), the definition of allostery may need to be broadened
16
The wedge-shaped protein monomers (red) assemble into trimers, but the alternative conformation for the monomer (square, green) forms tetramers The substrate or allosteric reg-ulator (yellow) binds only to the square conformation, and its binding prevents the square from adopting the wedge confor-mation Thus, if S or the allosteric regulator is present, equilib-rium favors a greater population of square tetramers among the morpheein ensemble at the expense of round trimers
17
18 Negative cooperativity in NADbinding to glyceraldehyde-3-P dehydrogenase:
0.2 0.4
0.6
Vmax v
0.8 1.0
3
Glutamine ATP or CTP
K0.5
[S]
S
S
Where affinity for NADH is
NAD
NAD
NAD Induced
conformation change
Trang 519 Hyperventilation results in decreased blood pCO2, which in turn
leads to an increase in pH Thus, the affinity of Hb for O2
in-creases (less O2will be released to tissues) Hypoventilation has
exactly the opposite effects on pCO2and pH; thus,
hypoventila-tion leads to a diminished affinity of Hb for O2(more O2will be
released to the tissues)
20 When hormone disappears, the hormone⬊receptor complex
dis-sociates, the GTPase activity of the Gsubunit cleaves bound
GTP to GDP Pi, and the affinity of G–GDP for adenylyl
cyclase is low, so it dissociates and adenylyl cyclase is no longer
activated Residual cAMP will be converted to 5-AMP by
phos-phodiesterase, and the catalytic subunits of protein kinase A will
be bound again by the regulatory subunits, whose cAMP ligands
have dissociated When protein kinase A becomes inactive,
phos-phorylase kinase will revert to the unphosphorylated, inactive
form through loss of phosphoryl groups Phophoprotein
phos-phatase 1 will act on glycogen phosphorylase a, removing the
phosphoryl group from Ser14and thereby converting glycogen
phosphorylase to the less active, allosterically regulated b form.
Chapter 16
1 The pronghorn antelope is truly a remarkable animal, with
nu-merous specially evolved anatomical and molecular features
These include a large windpipe (to draw in more oxygen and
exhale more carbon dioxide), lungs that are three times the size
of those of comparable animals (such as goats), and lung alveoli
with five times the surface area so that oxygen can diffuse more
rapidly into the capillaries The blood contains larger numbers
of red blood cells and thus more hemoglobin The skeletal and
heart muscles are likewise adapted for speed and endurance
The heart is three times the size of that of comparable animals
and pumps a proportionally larger volume of blood per
contrac-tion Significantly, the muscles contain much larger numbers of
energy-producing mitochondria, and the muscle fibers
them-selves are shorter and thus designed for faster contractions All
these characteristics enable the pronghorn antelope to run at a
speed nearly twice the top speed of a thoroughbred racehorse
and to sustain such speed for up to 1 hour
2 Refer to Figure 16.9 The step in which the myosin head
confor-mation change occurs, is the step that should be blocked by
,-methylene-ATP, because hydrolysis of ATP should occur in
this step and ,-methylene-ATP is nonhydrolyzable
3 Phosphocreatine is synthesized from creatine (via creatine
kinase) primarily in muscle mitochondria (where ATP is readily
generated) and then transported to the sarcoplasm, where it
can act as an ATP buffer The creatine kinase reaction in the
sar-coplasm yields the product creatine, which is transported back
into the mitochondria to complete the cycle Like many
mito-chondrial proteins, the expression of mitomito-chondrial creatine
kinase is directed by mitochondrial DNA, whereas sarcoplasmic
creatine kinase is derived from information encoded in nuclear
DNA
4 Note in step 3 of Figure 16.9 that it is ATP that stimulates
disso-ciation of myosin heads from the actin filaments—the
dissocia-tion of the cross-bridge complex When ATP levels decline (as
happens rapidly after death), large numbers of myosin heads
are unable to dissociate from actin and the muscle becomes stiff
and unable to relax
5 The skeletal muscles of the average adult male have a
cross-sectional area of approximately 35,000 cm2 The gluteus
maximus muscles represent approximately 300 cm2of this total
Assuming 4 kg of maximal tension per square centimeter of
cross-sectional area, one calculates a total tension for the gluteus maximus of 1200 kg (as stated in the problem) The same calcu-lation shows that the total tension that could be developed by all the muscles in the body is 140,000 kg (or 154 tons)!
6 Taking 55,000 g/mol divided by 6.02 1023/mol and by 1.3 g/mL, one obtains a volume of 7.03 1020mL Assume a sphere and use the volume of a sphere (V (4/3) r3) to obtain a radius of 25.6 Å The diameter of the tubulin dimer (see Figure 16.12) is 8 nm, or 80 Å, and two times the radius we calculated here is approximately 51 Å, a reasonable value by comparison
7 A liver cell is 20,000 nm long, which would correspond to 5000 tubulin monomers if using the value of Figure 16.12, and about
7800 tubulin monomers using the value of 25.6 Å calculated in problem 6
8 4 inches (length of the giant axon) 10.16 cm Movement at
2 to 5 traverse this distance
9 14 nm is 140 Å, which would be approximately 93 residues of a coiled coil
10 Using Equation 9.1, one can calculate a G of 17,100 J/mol for
this calcium gradient
11 Using Equation 3.13, one can calculate a cellular G of
48,600 J/mol for ATP hydrolysis
12 17,100 J are required to transport 1 mole of calcium ions Two moles of calcium would require 34,200 J, and three would cost 51,300 J Thus, the gradient of problem 10 would provide enough energy to drive the transport of two calcium ions per ATP hydrolyzed
13 Energy (or work) force distance If 1 ATP is hydrolyzed per step and the cellular value of G° for ATP hydrolysis is
50 kJ/mol, then the calculation of force exerted by a motor is force 50 kJ/mol/(step size) The SI unit of force is the newton, and 1 newton meter 1 J
For the kinesin-1 motor, 50 kJ/mol/(8 109m) 6.25
1012newtons
For the myosin-V motor, 50 kJ/mol/(36 109m) 1.39
1012newtons
For the dynein motor, 50 kJ/mol/(28 109m) 1.78
1012newtons
14 Perhaps the simplest way to view this problem is to consider the potential energy change for lifting a 10-kg weight 0.4 m
E mgh (10 kg)(9.8 m/sec2)(0.4 m) 39.2 J Now, if
1 ATP is expended per myosin step along an actin filament, 39.2 J/(50 kJ/mol) 7.8 104mol
Then, (7.8 104mol)(6.02 1023) 4.72 1020molecules
of ATP expended, one per step So there must be 4.72 1020
myosin steps along actin However, the stepping process is almost certainly not 100% efficient As shown in the box on page 489, a step size of 11 nm against a force of 4 pN would cor-respond to an energy expended per myosin per step of 4.4
1020J With this assumed energy expended per step, we can calculate 39 J/(4.4 1020J) 8.86 1020steps total
If we take the step size value for skeletal myosin from the box
on page 489 as 11 nm, one myosin head would have to take 0.4 m/11 109m or 3.64 107steps to raise the weight a dis-tance of 0.4 m Taking (8.86 1020steps total)/(3.64 107
steps per myosin) 2.43 1013myosin heads involved per 11-nm step
Trang 615 All these are smooth muscle except for the diaphragm (See
http://chanteur.net/contribu/index.htm#http://chanteur.net/contribu/
cJMdiaph.htm for an explanation of the common misconception
that the diaphragm is smooth muscle.)
16 1, a; 2, e; 3, d; 4, c; 5, b
17 This exercise is left to the student and will presumably be
differ-ent for every studdiffer-ent
18 The 80 Å step of the kinesin motor is a 0.008
10.16 cm would require 12.7 million steps
19 The correct answer is d, although each answer is reasonable
ATP is needed for several processes involved in muscle
relax-ation Salt imbalances can also prevent normal muscle function,
and interrupted blood flow could prevent efficient delivery of
oxygen needed for ATP production during cellular respiration
20 The correct answer is a Understanding the inheritance pattern
of sex-linked traits is essential Males always inherit sex-linked
traits (on the X chromosome) from their mothers In addition,
sex-linked traits are much more commonly expressed in males
because they have only one X chromosome
Chapter 17
1 6.5 1012(6.5 trillion) people
2 Consult Table 17.2
3 O2, H2O, and CO2
4 See Section 17.2
5 Consult Figure 17.8
6 Consult Corresponding Pathways of Catabolism and Anabolism Differ
in Important Ways, p 520 See also Figure 17.8.
7 See The ATP Cycle, p 521; NAD Collects Electrons Released in
Ca-tabolism, p 522; and NADPH Provides the Reducing Power for
Ana-bolic Processes, p 523.
8 In terms of quickness of response, the order is allosteric
regula-tion covalent modification enzyme synthesis and
degrada-tion See The Student Solutions Manual, Study Guide and Problems
Book for further discussions.
9 See Metabolic Pathways Are Compartmentalized within Cells, p 527,
and discussions in The Student Solutions Manual, Study Guide and
Problems Book.
10 Many answers are possible here Some examples: Large numbers
of metabolites in tissues and fluids; great diversity of structure of
biomolecules; need for analytical methods that can detect and
distinguish many different metabolites, often at very low
concen-trations; need to understand relationships between certain
metabolites; time and cost required to analyze and quantitate
many metabolites in a tissue or fluid
11 Mass spectrometry provides great sensitivity for detection of
metabolites NMR offers a variety of methods for resolving and
discriminating metabolites in complex mixtures Other
compar-isons and contrasts are discussed in the references at the end of
the chapter
12 See Figure 13.23; The Coenzymes of the Pyruvate
Dehydroge-nase Complex, page 568, and the Activation of Vitamin B12,
page 712
13 The genome is the entire hereditary information in an organism,
as encoded in its DNA or (for some viruses) RNA The
transcrip-tome is the set of all messenger RNA molecules (transcripts)
pro-duced in one cell or a population of cells under a defined set of
conditions The proteome is the entire complement of proteins produced by a genome, cell, tissue, or organism under a defined set of conditions The metabolome is the complete set of low molecular weight metabolites present in an organism or excreted
by it under a given set of physiological conditions
14 A mechanism for liver alcohol dehydrogenase:
15 TCA cycle: in mitochondria; converts acetate units to CO2plus NADH and FADH2
Glycolysis: in cytosol; converts glucose to pyruvate
Oxidative phosphorylation: in mitochondria; uses electrons (from NADH and FADH2) to produce ATP
Fatty acid synthesis: in cytosol; uses acetate units to synthesize fatty acids
16 Most discussions of ocean sequestration of CO2are devoted to capture of CO2by phytoplankton or injection of CO2deep in the ocean However, the box on page 512 offers another possibility—carbon sequestration in the shells of corals, mol-lusks, and crustaceans This approach may be feasible However, there are indications that these classes of sea organisms are already suffering from global warming and from pollution of ocean water
17 32P and 35S both decay via beta particle emission A beta particle
is merely an electron emitted by a neutron in the nucleus Thus, beta decay does not affect the atomic mass, but it does convert a neutron to a proton Thus, beta emission changes 32P to 32S and converts35S to 35Cl:
32P⎯⎯→ 32S, t1/2 14.3 days
35S⎯⎯→ 35Cl, t1/2 87.1 days
The decay equation is a first-order decay equation:
A/A0 e 0.693/t y2
Thus, after 100 days of decay, the fraction of 32P remaining would be 0.786% and the fraction of 35S remaining would
be 45%
18 The correct answer is b Obligate anaerobes can survive only in the absence of oxygen
19 The correct answer is a Fiber provides little or no nutrition The foods containing fiber, however, have other nutritional sub-stances that we are able to digest and absorb
Chapter 18
1 a Phosphoglucoisomerase, fructose bisphosphate aldolase, triose phosphate isomerase, glyceraldehyde-3-P dehydroge-nase, phosphoglycerate mutase, and enolase
b Hexokinase/glucokinase, phosphofructokinase, phospho-glycerate kinase, pyruvate kinase, and lactate dehydrogenase
c Hexokinase and phosphofructokinase
d Phosphoglycerate kinase and pyruvate kinase
P
C O
O O OC H H
H O
NH2
H A A
A
A
A
A A
B
N
C
O H H
NH2
A A
A A
A N
Trang 7e According to Equation 3.13, reactions in which the number
of reactant molecules differs from the number of product
molecules exhibit a strong dependence on concentration
That is, such reactions are extremely sensitive to changes
in concentration Using this criterion, we can predict from
Table 19.1 that the free energy changes of the fructose
bis-phosphate aldolase and glyceraldehyde-3-P dehydrogenase
reactions will be strongly influenced by changes in
concentration
f Reactions that occur with G near zero operate at or near
equilibrium See Table 18.1 and Figure 18.22
2 The carboxyl carbon of pyruvate derives from carbons 3 and
4 of glucose The keto carbon of pyruvate derives from carbons
2 and 5 of glucose The methyl carbon of pyruvate is obtained
from carbons 1 and 6 of glucose
3 Increased [ATP], [citrate], or [glucose-6-phosphate] inhibits
glycolysis Increased [AMP], [fructose-1,6-bisphosphate], or
[fructose-2,6-bisphosphate] stimulates glycolysis
4 See The Student Solutions Manual, Study Guide and Problems Book
for discussion
5 The mechanisms for fructose bisphosphate aldolase and
glyceraldehyde-3-P dehydrogenase are shown in Figures 18.12
and 18.13, respectively
6 The relevant reactions of galactose metabolism are shown in
Figure 18.24 See The Student Solutions Manual, Study Guide and
Problems Book for mechanisms.
7 Iodoacetic acid would be expected to alkylate the reactive
active-site cysteine that is vital to the glyceraldehyde-3-P dehydrogenase
reaction This alkylation would irreversibly inactivate the
enzyme
8 Ignoring the possibility that 32P might be incorporated into ATP,
the only reaction of glycolysis that utilizes Piis
glyceraldehyde-3-P dehydrogenase, which converts glyceraldehyde-glyceraldehyde-3-P to
1,3-bisphosphoglycerate.32Piwould label the phosphate at
carbon 1 of 1,3-bisphosphoglycerate The label will be lost in
the next reaction, and no other glycolytic intermediates will be
directly labeled (Once the label is incorporated into ATP, it
will also show up in glucose-6-P, fructose-6-P, and fructose-1,
6-bisphosphate.)
9 The sucrose phosphorylase reaction leads to glucose-6-P without
the need for the hexokinase reaction Direct production offers
the obvious advantage of saving a molecule of ATP
10 All of the kinases involved in glycolysis, as well as enolase, are
activated by Mg2ion A Mg2deficiency could lead to reduced
activity for some or all of these enzymes However, other
sys-temic effects of a Mg2deficiency might cause even more
serious problems
11 a 7.5
b 0.0266 mM.
12 1.08 kJ/mol
13 a 13.8 kJ/mol
b Keq 194,850
c [Pyr]/[PEP] 24,356
14 a 13.8 kJ/mol
b Keq 211
c [FBP]/[F-6-P] 528
15 a 19.1 kJ/mol
b Keq 1651
c [ATP]/[ADP] 13.8
16 G° 33.9 kJ/mol.
17 An 8% increase in [ATP] changes the [AMP] concentration to 20
18
Formation of Schiff base intermediate
Borohydride reduction
of Schiff base intermediate
Stable (trapped) E–S derivative
Degradation of enzyme (acid hydrolysis)
N6-dihydroxypropyl-L-lysine
PO3–
PO3– –
PO3–
CH2
CH2
O
B
Lys
Lys
OH
CH2
CH2
O
H H
H H
H
OH +
+
Lys
CH2
CH2
H
O H OH
Lys
CH2
CH2
H
H OH
OH
H2N
Trang 820 Pyruvate kinase deficiency primarily affects red blood cells,
which lack mitochondria and can produce ATP only via
glycoly-sis Absence of pyruvate kinase reduces the production of ATP
by glycolysis in red cells, which in turn reduces the activity of the
Na,K-ATPase (see Chapter 9) This results in alterations of
elec-trolyte (Naand K) concentrations, which leads to cellular
dis-tortion, rigidity, and dehydration Compromised red blood cells
are destroyed by the spleen and liver
21 The best answer is d PFK is more active at low ATP than at high
ATP, and F-2,6-bisP activates the enzyme
22 Hexokinase is inhibited by high concentrations of glucose-6-P,
so glycolysis would probably stop at such a high level of
glucose-6-P Also, the increased concentration of glucose-6-P would
change the cellular G of the hexokinase reaction to
approxi-mately22,000 J/mol
Chapter 19
1 Glutamate enters the TCA cycle via a transamination to form
-ketoglutarate The -carbon of glutamate entering the cycle is
equivalent to the methyl carbon of an entering acetate Thus,
no radioactivity from such a label would be lost in the first or
second cycle, but in each subsequent cycle, 50% of the total
label would be lost (see Figure 19.15)
2 NADactivates pyruvate dehydrogenase and isocitrate
dehydro-genase and thus would increase TCA cycle activity If NAD
increases at the expense of NADH, the resulting decrease in
NADH would likewise activate the cycle by stimulating citrate
synthase and -ketoglutarate dehydrogenase ATP inhibits
pyru-vate dehydrogenase, citrate synthase, and isocitrate
dehydroge-nase; reducing the ATP concentration would thus activate the
cycle Isocitrate is not a regulator of the cycle, but increasing its
concentration would mimic an increase in acetate flux through
the cycle and increase overall cycle activity
3 For most enzymes that are regulated by phosphorylation, the
co-valent binding of a phosphate group at a distant site induces a
conformation change at the active site that either activates or
in-hibits the enzyme activity On the other hand, X-ray
crystallo-graphic studies reveal that the phosphorylated and
unphosphorylated forms of isocitrate dehydrogenase share
iden-tical structures with only small (and probably insignificant)
con-formation changes at Ser113, the locus of phosphorylation What
phosphorylation does do is block isocitrate binding (with no
effect on the binding affinity of NADP) As shown in Figure 2
of the paper cited in the problem (Barford, D., 1991 Molecular
mechanisms for the control of enzymic activity by protein
phos-phorylation Biochimica et Biophysica Acta 1133:55–62), the
-carboxyl group of bound isocitrate forms a hydrogen bond
with the hydroxyl group of Ser113 Phosphorylation apparently
prevents isocitrate binding by a combination of a loss of the
crucial H bond between substrate and enzyme and by repulsive
electrostatic and steric effects
4 A mechanism for the first step of the -ketoglutarate
dehydroge-nase reaction:
5 Aconitase is inhibited by fluorocitrate, the product of citrate syn-thase action on fluoroacetate In a tissue where inhibition has occurred, all TCA cycle metabolites should be reduced in con-centration Fluorocitrate would replace citrate, and the concen-trations of isocitrate and all subsequent metabolites would be reduced because of aconitase inhibition
6 FADH2is colorless, but FAD is yellow, with a maximal ab-sorbance at 450 nm Succinate dehydrogenase could be conve-niently assayed by measuring the decrease in absorbance in a solution of the flavoenzyme and succinate
7 The central (C-3) carbon of citrate is reduced, and an adjacent carbon is oxidized in the aconitase reaction The carbon bearing the hydroxyl group is obviously oxidized by isocitrate dehydrogenase In the -ketoglutarate dehydrogenase reaction, the departing carbon atom and the carbon adjacent to it are both oxidized Both of the OCH2O carbons of succinate are oxidized in the succinate dehydrogenase reaction Of these four molecules, all but citrate undergo a net oxidation in the TCA cycle
8 Several TCA metabolite analogs are known, including malonate,
an analog of succinate, and 3-nitro-2-S-hydroxypropionate, the anion of which is a transition-state analog for the fumarase reaction
OH
A COO
A
CH2 COO
O
C
H
H
H E
[
(ð
H
OOC
H
B
OH
A C
E (ð E
H
OOC
O B O
H
O
N
Malonate (succinate analog)
3-Nitro-2-S-hydroxypropionate
Transition-state analog for fumarase
S
N
S
S
N
O
P
CH2 COO
C OH
TPP
A
O A
A
O
CH2
A
O P
CH2 COO
A
CH2
COO
A O
CH2 COO
CH2
N
N
HB
A
A O
CH2 COO
C OH
CH2
H H
O
B H
H
-KG
Covalent TPP intermediate
þ
k1 k–1
k–2
k2
E-UMP • Glc-1-P
Glc-1-P
k3 k–3
E-UMP
Gal-1-P
k4 k–4
E-UMP • Gal-1-P
k–5
k5
E-UDPGal UDPGal
k6 k–6
Trang 99 A mechanism for pyruvate decarboxylase:
12.14C incorporated in a reversed TCA cycle would label the two carboxyl carbons of oxaloacetate in the first pass through the cycle One of these would be eliminated in its second pass through the cycle The other would persist for more than two cycles and would be eliminated slowly as methyl carbons in acetyl-CoA in the reversed citrate synthase reaction
13 The labeling pattern would be the same as if methyl-labeled acetyl-CoA was fed to the conventional TCA cycle (see Figure 19.15)
C
CH3
COO–
O S
N R"
R
R'
B +
S N R"
R
R'
S N R"
R
R' + C
CH3
CO2
C OH
S N R"
R
R' + C
CH3 OH
S N R"
R
R'
C
CH3
OH
Resonance-stabilized carbanion on substrate
H+
S N R"
R
R' + C
CH3
H
Hydroxyethyl-TPP
S N R"
R
R' + –
O
E
E
H
Acetaldehyde
Pyruvate
– –
–
10 [isocitrate]/[citrate] 0.1 When [isocitrate] 0.03 mM,
[citrate] 0.3 mM.
11.14CO2incorporation into TCA via the pyruvate carboxylase
reac-tion would label the OCH2OCOOH carboxyl carbon in
oxalo-acetate When this entered the TCA cycle, the labeled carbon
would survive only to the -ketoglutarate dehydrogenase
reac-tion, where it would be eliminated as 14CO2
14
Acetyl-CoA
C
H2C
H
SCoA O
O COO–
Malate
C
O HC
O COO–
C
CoASH
O
OH
H2C
H2O
COO–
–
Trang 10Pyruvate
Cytosol
Malate
Oxaloacetate
Acetyl-CoA +
Citrate
ATP–Citrate lyase
Mitochondrial membrane
Malate
Oxaloacetate
Citrate
Acetyl-CoA +
Citrate synthase
Malate dehydrogenase
Malate dehydrogenase
Pyruvate
Mitochondrion
ATP
NADH
NADPH
NADH
NADP+
NAD+ NAD+
CoA
ADP Pi
CoA
CO 2
H2O
16 For the malate dehydrogenase reaction,
Malate NAD34 oxaloacetate NADH H
with the value of G° being 30 kJ/mol Then
G° RT ln Keq
ln (x/4.4 103)
12.1 ln (x/4.4 103)
x RT ln Keq
x
Using the dimensions given in the problem, one can calculate
a mitochondrial volume of 1.57 1015L (0.024 106M)
(1.57 1015L)(6.02 1023molecules/mole) 14.4, or about
14 molecules of OAA in a mitochondrion (“pOAA” 7.62)
17 This exercise is left to the student Review of the calculation of
oxidation numbers should be a prerequisite for answering this
problem
18 This exercise is left to the student
19 2R, 3R -fluorocitrate is converted by aconitase to
2-fluoro-cis-aconitate This intermediate then rotates 180 degrees in the
active site Addition of hydroxide at the C4 position is followed
by double bond migration from C3-C4 to C2-C3, to form
4-hydroxy-trans-aconitate This product remains tightly bound
at the aconitase active site, inactivating the enzyme Studies by
Trauble et al have shown that the inhibitory product can be
dis-placed by a 106-fold excess of isocitrate
30,000 J/mol
2478 J/mol
[1]x
[20][2.2 104]
20 Only eight ATPs would be generated in the succinyl-CoA synthe-tase (and nucleoside diphosphate kinase) reaction of the TCA cycle itself
21 d is false Succinyl-CoA is an inhibitor of citrate synthase
Chapter 20
1 The cytochrome couple is the acceptor, and the donor is the (bound) FAD/FADH2couple, because the cytochrome couple has a higher (more positive) reduction potential
Ᏹo0.254 V0.02 V*
*This is a typical value for enzyme-bound [FAD]
G nᏲᏱo 43.4 kJ/mol.
2.Ᏹo 0.03 V; G 5790 J/mol.
3 This situation is analogous to that described in Section 3.3 The net result of the reduction of NADis that a proton is con-sumed The effect on the calculation of free energy change is similar to that described in Equation 3.26 Adding an appropri-ate term to Equation 20.12 yields:
Ᏹ Ᏹo RT ln [H] (RT/nᏲ ln ([ox]/[red])
4 Cyanide acts primarily via binding to cytochrome a3, and the
amount of cytochrome a3in the body is much lower than the amount of hemoglobin Nitrite anion is an effective antidote for cyanide poisoning because of its unique ability to oxidize ferro-hemoglobin to ferriferro-hemoglobin, a form of ferro-hemoglobin that
competes very effectively with cytochrome a3for cyanide The amount of ferrohemoglobin needed to neutralize an otherwise