樱花的花语-幼儿园大班教学反思

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chapter
Fatty Acid Catabolism
17
in TriacylglycerolsOn a per- carbon basis, where does the largest amount of biologically
available energy in triacylglycerols reside: in the fatty acid portions or the glycerol portion? Indicate
how knowledge of the chemical structure of triacylglycerols provides the answer.
AnswerThe fatty acids of triacylglycerols are hydrocarbons, with a single carboxyl group.
Glycerol, on the other hand, has an OOH group on each carbon and is thus much more highly
oxidized than a fatty acid. On oxidation, fatty acids therefore produce far more energy per
carbon than does glycerol. Triacylglycerols have an energy of oxidation more than twice that
of the same weight of carbohydrates or proteins.
Reserves in Adipose TissueTriacylglycerols, with their hydrocarbon- like fatty acids, have the
highest energy content of the major nutrients.
(a)If 15% of the body mass of a 70.0 kg adult consists of triacylglycerols, what is the total available
fuel reserve, in kilojoules and kilocalories, in the form of triacylglycerols? Recall that
1.00 kcal ϭ4.18 kJ.
(b)If the basal energy requirement is approximately 8,400 kJday (2,000 kcalday), how long could
this person survive if the oxidation of fatty acids stored as triacylglycerols were the only source
of energy?
(c)What would be the weight loss in pounds per day under such starvation conditions (1 lb
ϭ0.454 kg)?
Answer
(a)Given (in the text) that the energy value of stored triacylglycerol is 38 kJg, the available
fuel reserve is
(0.15)(70.0 ϫ10
3
g)(38 kJg) ϭ4.0 ϫ10
5
kJ
ϭ9.6 ϫ10
4
kcal
(b)At a rate of 8.4 ϫ10
3
kJday, the fuel supply would last
(4.0 ϫ10
5
kJ)(8.4 ϫ10
3
kJday) ϭ48 days
(c)If all the triacylglycerol is used over a 48-day period, this represents a rate of weight loss of
ᎏᎏ
(0.15)(70.0 kg)
48 days
ϭ0.22 kgday
or (0.22 kgday)(0.454 kglb) ϭ0.48 lbday
Reaction Steps in the Fatty Acid Oxidation Cycle and Citric Acid CycleCells often
use the same enzyme reaction pattern for analogous metabolic conversions. For example, the steps in
the oxidation of pyruvate to acetyl- CoA and of a-ketoglutarate to succinyl-CoA, although catalyzed by
S-199

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S-200Chapter 17Fatty Acid Catabolism
different enzymes, are very similar. The first stage of fatty acid oxidation follows a reaction sequence
closely resembling a sequence in the citric acid cycle. Use equations to show the analogous reaction
sequences in the two pathways.
AnswerThe first three reactions in the boxidation of fatty acyl–CoA molecules are
analogous to three reactions of the citric acid cycle.
The fatty acyl–CoA dehydrogenase reaction is analogous to the succinate dehydrogenase
reaction; both are FAD-requiring oxidations:
Succinate ϩFAD
88n
fumarate ϩFADH
2
Fatty acyl–CoA ϩFAD
88n
trans-⌬
2
-enoyl-CoA ϩFADH
2
The enoyl-CoA hydratase reaction is analogous to the fumarase reaction; both add water to an
olefinic bond:
Fumarate ϩH
2
O
88n
malate
trans-⌬
2
-Enoyl-CoA ϩH
2
O
88n
L
-b-hydroxyacyl- CoA
The b-hydroxyacyl-CoA dehydrogenase reaction is analogous to the malate dehydrogenase
reaction; both are NAD-requiring and act on b-hydroxyacyl compounds:
Malate ϩNAD
ϩ
88n
oxaloacetate ϩNADH
L
-b-Hydroxyacyl-CoA ϩNAD
ϩ
88n
b-ketoacyl-CoA ϩNADH
4.␤Oxidation: How Many Cycles?How many cycles of ␤oxidation are required for the complete
oxidation of activated oleic acid, 18:1(⌬
9
)?
Answer7 cycles; the last releases 2 acetyl-CoA.
try of the Acyl-CoA Synthetase ReactionFatty acids are converted to their coenzyme A
esters in a reversible reaction catalyzed by acyl-CoA synthetase:
O
R
COO
Ϫ
ϩ ATP ϩ CoA
R
CCoA
ϩ
AMP
ϩ
PP
i
(a)The enzyme-bound intermediate in this reaction has been identified as the mixed anhydride of
the fatty acid and adenosine monophosphate (AMP), acyl-AMP:
OO
RCOPOCH
2
O
Adenine
O
Ϫ
H
HH
H
OH OH
Write two equations corresponding to the two steps of the reaction catalyzed by acyl- CoA
synthetase.
(b)The acyl-CoA synthetase reaction is readily reversible, with an equilibrium constant near 1. How
can this reaction be made to favor formation of fatty acyl–CoA?
AnswerActivation of carboxyl groups by ATP could in theory be accomplished by three types
of reactions: the formation of acyl- phosphate ϩADP; of acyl-ADP ϩP
i
; or of acyl-AMP ϩPP
i
.
All these reactions are readily reversible. To create an activation reaction with a highly nega-
tive ⌬GЈЊ(effectively irreversible), the third type of reaction can be coupled to a pyrophos-
phatase reaction, as in the synthesis of fatty acyl–CoA molecules.

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Chapter 17Fatty Acid CatabolismS-201
(a)ROCOO
Ϫ
ϩATP
88n
acyl-AMP ϩPP
i
Acyl-AMP ϩCoA
88n
acyl-CoA ϩAMP
(b)Hydrolysis of PP
i
by an inorganic pyrophosphatase pulls the reaction in the direction of
fatty acyl–CoA formation.
ediates in Oleic Acid OxidationWhat is the structure of the partially oxidized fatty acyl
group that is formed when oleic acid, 18:1(⌬
9
), has undergone three cycles of ␤oxidation? What are
the next two steps in the continued oxidation of this intermediate?
AnswerAfter three rounds of ␤oxidation, the fatty acyl–CoA has been shortened by six
carbons with the removal of three acetyl-CoAs. The resulting 12-carbon intermediate is
cis-⌬
3
-dodecanoyl-CoA, with the double bond between the third and fourth carbons from the
carboxyl end of the chain. Before another round of ␤oxidation can occur, that double bond
must be moved from ⌬
3
to ⌬
2
, which is catalyzed by ⌬
3
,⌬
2
-enoyl isomerase (see Fig. 17–9).
Water is then added to the double bond to form the ␤-hydroxydodecanoyl-CoA derivative,
which can undergo further ␤oxidation.
7.␤Oxidation of an Odd-Chain Fatty AcidWhat are the direct products of ␤oxidation of a fully
saturated, straight-chain fatty acid of 11 carbons?
Answer4 acetyl-CoA and 1 propionyl- CoA
ion of Tritiated PalmitatePalmitate uniformly labeled with tritium (
3
H) to a specific activ-
ity of 2.48 ϫ10
8
counts per minute (cpm) per micromole of palmitate is added to a mitochondrial
preparation that oxidizes it to acetyl-CoA. The acetyl-CoA is isolated and hydrolyzed to acetate. The
specific activity of the isolated acetate is 1.00 ϫ10
7
cpmmmol. Is this result consistent with the
b-oxidation pathway? Explain. What is the final fate of the removed tritium?
AnswerThe b-oxidation pathway includes two dehydrogenase enzymes that remove
hydrogen (H–H) from the fatty acyl–CoA chain, first at a OCH
2
OCH
2
Oand then at a
OCH
2
OCH(OH)O. The net result of the two reactions is removal of one of the two hydro-
gens at the point of formation of the enoyl-CoA intermediate. The two other hydrogens in the
methyl group of acetyl-CoA come from water.
Palmitate contains 16 carbons, with (14 ϫ2) ϩ3 ϭ31 hydrogens, so each two-carbon unit
contains about 431 or about 18 of the total
3
H present. Thus, the counts per minute expected
per acetyl-CoA, with two of the four acetyl hydrogens labeled (the other two arising from unla-
beled water), is (24)(2.48 ϫ10
8
cpmmmol)(18) ϭ1.6 ϫ10
7
cpmmmol, somewhat higher
than observed. Exchange between b-ketoacyl-CoA and solvent water could cause loss of
3
H.
The final fate of the tritium removed from palmitate is its appearance in water, as reduced
carriers (FADH
2
, NADH) are reoxidized by the mitochondria.
tmentation in b OxidationFree palmitate is activated to its coenzyme A derivative
(palmitoyl-CoA) in the cytosol before it can be oxidized in the mitochondrion. If palmitate and
[
14
C]coenzyme A are added to a liver homogenate, palmitoyl-CoA isolated from the cytosolic fraction
is radioactive, but that isolated from the mitochondrial fraction is not. Explain.
AnswerThe transport of fatty acid molecules into mitochondria requires a shuttle system in-
volving a fatty acyl–carnitine intermediate. Fatty acids are first converted to fatty acyl–CoA
molecules in the cytosol (by the action of acyl–CoA synthetases) then, at the outer mitochon-
drial membrane, the fatty acyl group is transferred to carnitine (by the action of carnitine acyl-
transferase I). After transport of fatty acyl–carnitine through the inner membrane, the fatty
acyl group is transferred to mitochondrial CoA. The cytosolic and mitochondrial pools of CoA
are thus kept separate, and no labeled CoA from the cytosolic pool enters the mitochondrion.

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S-202Chapter 17Fatty Acid Catabolism
ative Biochemistry: Energy- Generating Pathways in BirdsOne indication of the rela-
tive importance of various ATP-producing pathways is the V
max
of certain enzymes of these pathways.
The values of V
max
of several enzymes from the pectoral muscles (chest muscles used for flying) of
pigeon and pheasant are listed below.
V
max
(mmol substrateming tissue)
EnzymePigeonPheasant
H exokinase3.02.3
Glycogen phosphorylase18.0120.0
Phosphofructokinase-124.0143.0
Citrate synthase100.015.0
Triacylglycerol lipase0.070.01
(a)Discuss the relative importance of glycogen metabolism and fat metabolism in generating ATP in
the pectoral muscles of these birds.
(b)Compare oxygen consumption in the two birds.
(c)Judging from the data in the table, which bird is the long- distance flyer? Justify your answer.
(d)Why were these particular enzymes selected for comparison? Would the activities of triose phos-
phate isomerase and malate dehydrogenase be equally good bases for comparison? Explain.
Answer
(a)In the pigeon, aerobic oxidation of fatty acids—boxidation and oxidative phosphorylation—
predominates; in the pheasant, anaerobic glycolysis of glycogen predominates. Note the
high citrate synthase activity in the pigeon, and the high glycogen phosphorylase and
PFK-1 activities in the pheasant.
(b)Using aerobic oxidation, pigeon muscle consumes more oxygen during flight.
(c)The energy available per gram is higher for fat than for glycogen. In addition, anaerobic
breakdown of glycogen is limited by tolerance to lactate buildup. Thus the pigeon, using
predominantly the oxidative catabolism of fats, is the long-distance flyer.
(d)The enzymes listed in the table (unlike triose phosphate isomerase and malate dehydro-
genase) are the regulatory enzymes of their respective pathways and thus limit ATP
production rates.
Carnitine AcyltransferaseWhat changes in metabolic pattern would result from a muta-
tion in the muscle carnitine acyltransferase I in which the mutant protein has lost its affinity for
malonyl-CoA but not its catalytic activity?
AnswerMalonyl-CoA would no longer inhibit fatty acid entry into the mitochondrion and
␤oxidation, so there might be a futile cycle of simultaneous fatty acid synthesis in the cytosol
and fatty acid breakdown in mitochondria. (See Fig. 17–12.)
of Carnitine DeficiencyAn individual developed a condition characterized by progressive
muscular weakness and aching muscle cramps. The symptoms were aggravated by fasting, exercise,
and a high-fat diet. The homogenate of a skeletal muscle specimen from the patient oxidized added
oleate more slowly than did control homogenates, consisting of muscle specimens from healthy indi-
viduals. When carnitine was added to the patient’s muscle homogenate, the rate of oleate oxidation
equaled that in the control homogenates. The patient was diagnosed as having a carnitine deficiency.
(a)Why did added carnitine increase the rate of oleate oxidation in the patient’s muscle homogenate?

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Chapter 17Fatty Acid CatabolismS-203
(b)Why were the patient’s symptoms aggravated by fasting, exercise, and a high-fat diet?
(c)Suggest two possible reasons for the deficiency of muscle carnitine in this individual.
Answer
(a)The carnitine-mediated transport of fatty acids into mitochondria is the rate-limiting
step in boxidation (see Fig. 17–6). Carnitine deficiency decreases the rate of transport
of fatty acids into mitochondria and thus the rate of boxidation, so addition of carnitine
would increase the rate of oxidation.
(b)Fasting, exercise, and a high-fat diet all cause an increased need for boxidation of fatty
acids and thus an increased demand for carnitine shuttle activity. The symptoms of car-
nitine deficiency would therefore become more severe under these conditions.
(c)The deficiency of carnitine may result from a dietary deficiency of its precursor, lysine,
or from a defect in one of the enzymes that synthesize carnitine from this precursor.
Fatty Acids as a Source of WaterContrary to legend, camels do not store water in their humps,
which actually consist of large fat deposits. How can these fat deposits serve as a source of water?
Calculate the amount of water (in liters) that a camel can produce from 1.0 kg of fat. Assume for
simplicity that the fat consists entirely of tripalmitoylglycerol.
AnswerOxidation of fatty acids produces water in significant amounts. From Equation 17–6
Palmitoyl-CoA ϩ23O
2
ϩ108P
i
ϩ108ADP
88n
CoA ϩ16CO
2
ϩ108ATP ϩ23H
2
O
we know that the oxidation of 1 mol of palmitoyl-CoA produces 23 mol of water.
Tripalmitoin (glycerol plus three palmitates in ester linkage) has a molecular weight of 885, so
1 kg of tripalmitoin contains (1.0 kg)(1,000 gkg)(885 gmol) ϭ1.1 mol. Complete oxidation
of the three palmitoyl groups will produce
(1.1moltripalmit oin)(3molpalmitatemoltripalmitoin)(23 molH
2
Omol palmitate)ϭ76 molH
2
O
Thus, the volume of water produced (ignoring the contribution of glycerol oxidation) is
(76 mol)(18 gmol)(1 kg1,000 g)(1 Lkg) ϭ1.4 L
Note:in reality, this may be an overestimate. The fatty acyl groups of the triacylglycerol in
the camel’s fat may be less highly reduced than palmitate.
Petroleum as a Microbial Food SourceSome microorganisms of the genera Nocardiaand
Pseudomonascan grow in an environment where hydrocarbons are the only food source. These
bacteria oxidize straight-chain aliphatic hydrocarbons, such as octane, to their corresponding
carboxylic acids:
CH
3
( CH
2
)
6
CH
3
ϩNAD
ϩ
ϩO
2
88z
y88CH
3
(CH
2
)
6
C OOH ϩNADH ϩH
ϩ
How could these bacteria be used to clean up oil spills? What would be some of the limiting factors to
the efficiency of this process?
AnswerBy oxidizing hydrocarbons to their corresponding fatty acids, these microbes can ob-
tain all their energy from boxidation and oxidative phosphorylation, converting the hydrocar-
bons to CO
2
and H
2
O. Theoretically, oil spills could be broken down by treatment with these
microbes.
Because of the extreme hydrophobicity of hydrocarbons, close contact between substrate
and bacterial enzymes might be difficult to achieve; under field conditions (e.g., an oil spill),
detergents are often added to improve this contact. In addition, other nutrients, such as nitro-
gen or phosphorus, may be limiting for the bacterial populations, and these elements are often
added to foster the growth of the hydrocarbon-oxidizers.
13.
14.

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S-204Chapter 17Fatty Acid Catabolism
lism of a Straight-Chain Phenylated Fatty AcidA crystalline metabolite was isolated from
the urine of a rabbit that had been fed a straight-chain fatty acid containing a terminal phenyl group:
CH
2
(CH
2
)
n
COO
Ϫ
A 302 mg sample of the metabolite in aqueous solution was completely neutralized by 22.2 mL of
0.100
M
NaOH.
(a)What is the probable molecular weight and structure of the metabolite?
(b)Did the straight-chain fatty acid contain an even or an odd number of methylene (OCH
2
O)
groups (i.e., is neven or odd)? Explain.
Answer
(a)22.2 mL of 0.1
M
NaOH is equivalent to (22.2 ϫ10
Ϫ3
L) (0.100 molL) ϭ22.2 ϭ10
Ϫ4
mol
of unknown metabolite (assuming that it contains only one carboxyl group) in the 302 mg
sample. Thus, the M
r
of the metabolite is
ᎏᎏ
302 ϫ10
Ϫ3
g
22.2 ϫ10
Ϫ4
mol
ϭ136
This is the M
r
of phenylacetic acid.
(b)Because boxidation removes two-carbon units, and the end product is a two-carbon
unit, the original fatty acyl chain must have had an even number of methylene groups
(with the phenyl group counted as equivalent to a terminal methyl group). An odd-
numbered fatty acid would have produced phenylpropionate.
Acid Oxidation in Uncontrolled DiabetesWhen the acetyl-CoA produced during boxida-
tion in the liver exceeds the capacity of the citric acid cycle, the excess acetyl-CoA forms ketone
bodies—acetone, acetoacetate, and
D
-b-hydroxybutyrate. This occurs in severe, uncontrolled diabetes:
because the tissues cannot use glucose, they oxidize large amounts of fatty acids instead. Although
acetyl-CoA is not toxic, the mitochondrion must divert the acetyl-CoA to ketone bodies. What problem
would arise if acetyl-CoA were not converted to ketone bodies? How does the diversion to ketone
bodies solve the problem?
AnswerIndividuals with uncontrolled diabetes oxidize large quantities of fat because they
cannot use glucose efficiently. This leads to a decrease in activity of the citric acid cycle (see
Problem 17) and an increase in the pool of acetyl-CoA. If acetyl-CoA were not converted to
ketone bodies, the CoA pool would become depleted. Because the mitochondrial CoA pool
is small, liver mitochondria recycle CoA by condensing two acetyl-CoA molecules to form
acetoacetyl-CoA ϩCoA (see Fig. 17–18). The acetoacetyl-CoA is converted to other ketones,
and the CoA is recycled for use in the b-oxidation pathway and energy production.
uences of a High-Fat Diet with No CarbohydratesSuppose you had to subsist on a diet
of whale blubber and seal blubber, with little or no carbohydrate.
(a)What would be the effect of carbohydrate deprivation on the utilization of fats for energy?
(b)If your diet were totally devoid of carbohydrate, would it be better to consume odd- or even-
numbered fatty acids? Explain.
Answer
(a)Pyruvate, formed from glucose via glycolysis, is the main source of the oxaloacetate
needed to replenish citric acid cycle intermediates (see Table 16–2 ). In the absence of
carbohydrate in the diet, the oxaloacetate level drops and the citric acid cycle slows.
This increases the rate of boxidation of fatty acids and leads to ketosis.

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Chapter 17Fatty Acid CatabolismS-205
(b)The last cycle of boxidation produces two acetyl-CoA molecules from an even-
numbered fatty acid, or propionyl-CoA ϩacetyl-CoA from an odd-numbered fatty acid.
Propionyl-CoA can be converted to succinyl-CoA (see Fig. 17–11), which when con-
verted to oxaloacetate stimulates the citric acid cycle and relieves the conditions leading
to ketosis. Thus, it would be better to consume odd-numbered fatty acids.
- and Odd- Chain Fatty Acids in the DietIn a laboratory experiment, two groups of rats are
fed two different fatty acids as their sole source of carbon for a month. The first group gets heptanoic
acid (7:0), and the second gets octanoic acid (8:0). After the experiment, a striking difference is seen
between the two groups. Those in the first group are healthy and have gained weight, whereas those
in the second group are weak and have lost weight as a result of losing muscle mass. What is the
biochemical basis for this difference?
AnswerThe ␤oxidation of heptanoic acid (which has an odd number of carbons) produces
the three-carbon intermediate propionyl-CoA, which can be converted by propionyl-CoA car-
boxylase to methylmalonyl- CoA, then to succinyl-CoA. This four-carbon product of fatty acid
oxidation can then be converted to oxaloacetate in the citric acid cycle, and the oxaloacetate
can be used for gluconeogenesis—thus providing the animal with carbohydrate as well as en-
ergy from fatty acid oxidation. Animals fed octanoic acid (with an even number of carbons)
degrade it completely to acetyl-CoA by three rounds of ␤oxidation. This provides energy via
the citric acid cycle but does not provide starting material for gluconeogenesis. These animals
are therefore deficient in glucose, the primary fuel for the brain and an intermediate in many
biosynthetic pathways.
lic Consequences of Ingesting q-FluorooleateThe shrub Dichapetalum toxicarium,
native to Sierra Leone, produces q-fluorooleate, which is highly toxic to warm- blooded animals.
H
H
F
CH
2
(CH< br>2
)
7
C
C
(CH
2
)
7
COO
Ϫ
␻-Fluorooleate
This substance has been used as an arrow poison, and powdered fruit from the plant is sometimes used
as a rat poison (hence the plant’s common name, ratsbane). Why is this substance so toxic? (Hint: re-
view Chapter 16, Problem 22.)
AnswerOxidation of q-fluorooleate in the b-oxidation pathway forms fluoroacetyl-CoA in the
last pass through the sequence. Entry of fluoroacetyl-CoA into the citric acid cycle produces
fluorocitrate, a powerful inhibitor of the enzyme aconitase. As a result of this inhibition, the
citric acid cycle shuts down and the flow of reducing equivalents to oxidative phosphorylation
is fatally impaired.
Acetyl-CoA CarboxylaseWhat would be the consequences for fat metabolism of a muta-
tion in acetyl-CoA carboxylase that replaced the Ser residue normally phosphorylated by AMPK to an
Ala residue? What might happen if the same Ser were replaced by Asp? (Hint: See Fig. 17–12.)
AnswerThe Ser-to-Ala change would produce an enzyme that could not be inhibited by
phosphorylation by AMPK. The first step in fatty acid synthesis would be constantly turned
on, and the malonyl-CoA produced by acetyl-CoA carboxylase would inhibit entry of fatty
acids into mitochondria, shutting down ␤oxidation. The Ser-to-Asp mutation would put a
negatively charged Asp residue in the position occupied by
P
-Ser in the inhibited wild-type
enzyme. This might mimic the effect of a phosphorylated Ser residue, shutting down
acetyl-CoA carboxylase, inhibiting fatty acid synthesis, and stimulating ␤oxidation.

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S-206Chapter 17Fatty Acid Catabolism
of PDE Inhibitor on AdipocytesHow would an adipocyte’s response to epinephrine be
affected by the addition of an inhibitor of cAMP phosphodiesterase (PDE)? (Hint: See Fig. 12–4.)
AnswerResponse to glucagon or epinephrine would be prolonged because cAMP, once
formed, would persist, stimulating protein kinase A for a longer period and leading to longer-
lasting mobilization of fatty acids in adipocytes.
of FAD as Electron AcceptorAcyl- CoA dehydrogenase uses enzyme-bound FAD as a pros-
thetic group to dehydrogenate the aand bcarbons of fatty acyl–CoA. What is the advantage of using
FAD as an electron acceptor rather than NAD
ϩ
? Explain in terms of the standard reduction potentials
for the Enz- FADFADH
2
(EЈЊϭϪ0.219 V) and NAD
ϩ
NADH (EЈЊϭϪ0.320 V) half- reactions.
AnswerEnz-FAD, having a more positive standard reduction potential, is a better electron
acceptor than NAD
ϩ
, and the reaction is driven in the direction of fatty acyl–CoA oxidation (a
negative free-energy change). This more favorable free-energy change is obtained at the ex-
pense of 1 ATP; only 1.5 ATP molecules are formed per FADH
2
oxidized in the respiratory
chain, compared with 2.5 ATP per NADH.
23.b Oxidation of Arachidic AcidHow many turns of the fatty acid oxidation cycle are required for
complete oxidation of arachidic acid (see Table 10–1) to acetyl- CoA?
AnswerArachidic acid is a 20-carbon saturated fatty acid. Nine cycles of the b-oxidation
pathway are required for its oxidation, producing 10 molecules of acetyl-CoA, the last two in
the ninth turn.
of Labeled PropionateIf [3-
14
C]propionate (
14
C in the methyl group) is added to a liver ho-
mogenate,
14
C-labeled oxaloacetate is rapidly produced. Draw a flow chart for the pathway by which
propionate is transformed to oxaloacetate, and indicate the location of the
14
C in oxaloacetate.
AnswerPropionate is first converted to the CoA derivative. Figure 17–11 shows the three-
step pathway that converts propionyl-CoA to succinyl-CoA, which can be summarized as fol-
lows. Use these descriptions to prepare your own flow diagram.
nyl-CoA carboxylase uses CO
2
and ATP to form
D
-methylmalonyl-CoA by carboxy-
lation at C-2 of the propionyl group.
malonyl-CoA epimerase shifts the CoA thioester from C-1 (of the original propi-
onyl group) to the newly added carboxylate, making the product
L
-methylmalonyl-CoA.
malonyl-CoA mutase moves the carboxy-CoA group from C-2 to C-3 within the
original propionyl unit, forming succinyl- CoA.
succinyl-CoA is formed, the citric acid cycle can convert it to oxaloacetate.
Given the stereochemistry of these reactions, the [
14
C]-label is equilibrated at C-2 and
C-3 of the oxaloacetate.
ic Acid MetabolismWhen phytanic acid uniformly labeled with
14
C is fed to a mouse, ra-
dioactivity can be detected in malate, a citric acid cycle intermediate, within minutes. Draw a meta-
bolic pathway that could account for this. Which of the carbon atoms in malate would contain
14
C
label?
AnswerPhytanic acid is degraded to pristanic acid by the pathway shown in Figure 17–17.
Pristanic acid undergoes ␤oxidation, with each round yielding propionyl-CoA (not acetyl-CoA,
as for a straight-chain fatty acid). Degradation of uniformly labeled phytanic acid produces

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Chapter 17Fatty Acid CatabolismS-207
propionyl-CoA labeled in all three carbons of propionate. Propionyl-CoA is converted to suc-
cinyl-CoA by the series of reactions shown in Figure 17–11. The C-2 and C-3 of the succinyl
moiety are labeled, and either C-1 or C-4 as well. When this succinate is converted to malate
in the citric acid cycle, the malate is labeled at C-2 and C-3, and labeled half as much at C-1
and C-4.
s of H
2
O Produced in bOxidationThe complete oxidation of palmitoyl-CoA to carbon
dioxide and water is represented by the overall equation
Palmitoyl-CoA ϩ23O
2
ϩ108P
i
ϩ108ADP
88n
CoA ϩ16CO
2
ϩ108ATP ϩ23H
2
O
Water is also produced in the reaction
ADP ϩP
i
88n
ATP ϩH
2
O
but is not included as a product in the overall equation. Why?
AnswerATP hydrolysis in the cell’s energy- requiring reactions uses water, in the reaction
ATP ϩH
2
O
88n
ADP ϩP
i
In a cell at steady state, for every mole of ATP hydrolyzed, a mole of ATP is formed by con-
densation of ADP ϩP
i
. There is no netchange in [ATP] and thus no netproduction of H
2
O.
ical Importance of CobaltIn cattle, deer, sheep, and other ruminant animals, large amounts
of propionate are produced in the rumen through the bacterial fermentation of ingested plant matter.
Propionate is the principal source of glucose for these animals, via the route propionate noxaloac-
etate nglucose. In some areas of the world, notably Australia, ruminant animals sometimes show
symptoms of anemia with concomitant loss of appetite and retarded growth, resulting from an inability
to transform propionate to oxaloacetate. This condition is due to a cobalt deficiency caused by very
low cobalt levels in the soil and thus in plant matter. Explain.
AnswerOne of the enzymes necessary for the conversion of propionate to oxaloacetate is
methylmalonyl-CoA mutase (see Fig. 17–11). This enzyme requires as an essential cofactor
the cobalt-containing coenzyme B
12
, which is synthesized from vitamin B
12
. A cobalt defi-
ciency in animals would result in coenzyme B
12
deficiency.
Loss during HibernationBears expend about 25 ϫ10
6
Jday during periods of hibernation,
which may last as long as seven months. The energy required to sustain life is obtained from fatty acid
oxidation. How much weight loss (in kilograms) has occurred after seven months? How might ketosis be
minimized during hibernation? (Assume the oxidation of fat yields 38 kJg.)
AnswerIf the catabolism of fat yields 38 kJg, or 3.8 ϫ10
4
kJkg, and the bear expends
25 ϫ10
6
Jday, or 2.5 ϫ10
4
kJday, then the bear will lose
(2.5 ϫ10
4
kJday)(3.8 ϫ10
4
kJkg) ϭ0.66 kgday
and in 7 months, or 210 days, will lose
0.66 kgday ϫ210 days ϭ140 kg
To minimize ketosis, a slow but steady degradation of nonessential proteins would provide
three-, four-, and five-carbon products essential to the formation of glucose by gluconeogene-
sis. This would avoid the inhibition of the citric acid cycle that occurs when oxaloacetate is
withdrawn from the cycle to be used for gluconeogenesis. The citric acid cycle could continue
to degrade acetyl-CoA, rather than shunting it into ketone body formation.

2608T_ch17sm_S199-S210 02222008 2:07 pm Page S-208 pinnacle 111:WHQY028:Solutions Manual:Ch-17:
S-208Chapter 17Fatty Acid Catabolism
Data Analysis Problem
29.␤Oxidation of Trans FatsUnsaturated fats with trans double bonds are commonly referred to as
“trans fats.” There has been much discussion about the effects of dietary trans fats on health. In their
investigations of the effects of trans fatty acid metabolism on health, Yu and colleagues (2004) showed
that a model trans fatty acid was processed differently from its cis isomer. They used three related
18-carbon fatty acids to explore the difference in ␤oxidation between cis and trans isomers of the
same-size fatty acid.
O
OH
(octadecenoic acid)
Stearic acid
O
OH
Oleic acid
(cis-⌬
9
-octadecenoic acid)
O
OH
(trans-⌬
Elaidic acid
9
-octadecenoic acid)
The researchers incubated the coenzyme A derivative of each acid with rat liver mitochondria for
5 minutes, then separated the remaining CoA derivatives in each mixture by HPLC (high- performance
liquid chromatography). The results are shown below, with separate panels for the three experiments.
+
Stearoyl-CoA
mitochondria+ mitochondria
Oleoyl- CoA
+
Elaidoyl-CoA
mitochondria
cΔ
9
C
18
-CoA
tΔ
9
C
18
-Co A
C
18
-CoA
tΔ
5
C
14
-CoA
m
IS
n

4
5
IS
2

t
a
IS

e
c
n
a
b
r
o
s
b
A
cΔ
5
C
14
-Co A
12
21
30
Time (min)
In the figure, IS indicates an internal standard (pentadecanoyl-CoA) added to the mixture, after
the reaction, as a molecular marker. The researchers abbreviated the CoA derivatives as follows:
stearoyl-CoA, C
18
-CoA; cis-⌬
5
-tetradecenoyl-CoA, c⌬
5
C
5
14
-CoA; oleoyl-CoA, c⌬
9
C
18
-CoA;
trans-⌬
5
-tetradecenoyl-CoA, t⌬C
14
-CoA; and elaidoyl-CoA, t⌬
9
C
18
- CoA.

2608T_ch17sm_S199-S210 02222008 2:07 pm Page S-209 pinnacle 111:WHQY028:Solutions Manual:Ch-17:
Chapter 17Fatty Acid CatabolismS- 209
O
S-CoA
cis-⌬
5
-Tetradecenoyl- CoA
O
S-CoA
trans-⌬
5
-Tetradecenoy l-CoA
(a)Why did Yu and colleagues need to use CoA derivatives rather than the free fatty acids in these
experiments?
(b)Why were no lower molecular weight CoA derivatives found in the reaction with stearoyl-CoA?
(c)How many rounds of ␤oxidation would be required toconvert the oleoyl-CoA and the elaidoyl-
CoA to cis-⌬
5
-tetradecenoyl-CoA and trans-⌬
5
-tetradecenoyl-CoA, respectively?
There are two forms of the enzyme acyl-CoA dehydrogenase (see Fig. 17–8a): long- chain
acyl-CoA dehydrogenase (LCAD) and very- long-chain acyl-CoA dehydrogenase (VLCAD). Yu and
coworkers measured the kinetic parameters of both enzymes. Theyused the CoA derivatives of
three fatty acids: tetradecanoyl-CoA (C
14
-CoA), cis-⌬
5
-tetradecenoyl- CoA (c⌬
5
C
14
-CoA), and
trans-⌬
5
-tetradecenoyl- CoA(t⌬
5
C
14
-CoA). The results are shown below. (See Chapter 6 for
definitions of the kinetic parameters.)
LCADVLCAD
C
14-c⌬
5
C
CoA
14
-t⌬
5
Cc⌬
5
C-t⌬
5
C
CoA
14
-C
CoACoA< br>14
-
CoA
14
CoA
14
-
V
max
3.33.02.91.40.320.88
K
m
0.410.401 .60.570.440.97
k
cat
9.98.98.52.00.421.12
k
cat
K
m
24225411
(d)For LCAD, the K
m
differs dramatically for the cis and trans substrates. Provide a plausible expla-
nation for this observation in terms of the structures of the substrate molecules. (Hint: You may
want to refer to Fig. 10–2.)
(e)The kinetic parameters of the two enzymes are relevant to the differential processing of these
fatty acids onlyif the LCAD or VLCAD reaction (or both) is the rate-limiting step in the path-
way. What evidence is there to support this assumption?
(f)How do these different kinetic parameters explain the different levels of the CoA derivatives
found after incubation of rat liver mitochondria with stearoyl-CoA, oleoyl-CoA, and elaidoyl-CoA
(shown in the three-panel figure)?
Yu and coworkers measured the substrate specificity of rat liver mitochondrial thioesterase, which
hydrolyzes acyl-CoA to CoA and free fatty acid (see Chapter 21). This enzyme was approximately
twice as active with C
14
-CoA thioesters as with C
18
-CoA thioesters.
(g)Other research has suggested that free fatty acids can pass through membranes. In their experi-
ments, Yu and colleagues found trans-⌬
5
-tetradecenoic acid outside mitochondria (i.e., in the
medium) that had been incubated with elaidoyl-CoA. Describe the pathway that led to this ex-
tramitochondrial trans-⌬
5
-tetradecenoic acid. Be sure to indicate where in the cell the various
transformations take place, as well as the enzymes that catalyze the transformations.

2608T_ch17sm_S199-S210 02222008 2:07 pm Page S-210 pinnacle 111:WHQY028:Solutions Manual:Ch-17:
S-210Chapter 17Fatty Acid Catabolism
(h)It is often said in the popular press that “trans fats are not broken down by your cells and instead
accumulate in your body.” In what sense is this statement correct and in what sense is it an over-
simplification?
Answer
(a)Fatty acids are converted to their CoA derivatives by enzymes in the cytoplasm; the
acyl-CoAs are then imported into mitochondria for oxidation. Given that the researchers
were using isolated mitochondria, they had to use CoA derivatives.
(b)Stearoyl-CoA was rapidly converted to 9 acetyl-CoA by the ␤-oxidation pathway. All in-
termediates reacted rapidly and none were detectable at significant levels.
(c)Two rounds. Each round removes two carbon atoms, thus two rounds convert an
18-carbon to a 14-carbon fatty acid and 2 acetyl-CoA.
(d)The K
m
is higher for the trans isomer than for the cis, so a higher concentration of trans
isomer is required for the same rate of breakdown. Roughly speaking, the trans isomer
binds less well than the cis, probably because differences in shape, even though not at
the target site for the enzyme, affect substrate binding to the enzyme.
(e)The substrate for LCADVLCAD builds up differently, depending on the particular sub-
strate; this is expected for the rate-limiting step in a pathway.
(f)The kinetic parameters show that the trans isomer is a poorer substrate than the cis for
LCAD, but there is little difference for VLCAD. Because it is a poorer substrate, the
trans isomer accumulates to higher levels than the cis.
(g)One possible pathway is shown below (indicating “inside” and “outside” mitochondria).
Elaidoyl-CoA
carnitine
elaidoyl-carnitine
carnitine
transport
elaidoyl-carnitineelaidoyl- CoA
2 rounds
acyltransferase I
acyltransferase II
of ␤ oxidation
(outs ide)(outside)
(inside)(inside)
5-trans- tetradecenoyl-CoA
thioesterase
5-trans- tetradecanoic acid
diffusion
5-trans- tetradecanoic acid
(inside)
(inside)
(outside)
(h)It is correct insofar as trans fats are broken down less efficiently than cis fats, and thus
trans fats may “leak” out of mitochondria. It is incorrect to say that trans fats are not
broken down by cells; they are broken down, but at a slower rate than cis fats.
Reference
Yu, W., Liang, X., Ensenauer, R., Vockley, J., Sweetman, L., & Schultz, H.(2004) Leaky ␤-oxidation of a trans-fatty acid.
J. . 279,52,160–52,167.