Resolução Lehninger - Capítulo 10

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S-112
Lipids
chapter 
10
1. Operational Definition of Lipids How is the definition of “lipid” different from the types of definitions
used for other biomolecules that we have considered, such as amino acids, nucleic acids, and proteins?
Answer The term “lipid” does not specify a particular chemical structure. Whereas one can
write a general formula for an amino acid, nucleic acid, or protein, lipids are much more
chemically diverse. Compounds are categorized as lipids based on their greater solubility in
organic solvents than in water.
2. Melting Points of Lipids The melting points of a series of 18-carbon fatty acids are: stearic acid, 
69.6 �C; oleic acid, 13.4 �C; linoleic acid, �5 �C; and linolenic acid, �11 �C.
(a) What structural aspect of these 18-carbon fatty acids can be correlated with the melting point?
(b) Draw all the possible triacylglycerols that can be constructed from glycerol, palmitic acid, and
oleic acid. Rank them in order of increasing melting point.
(c) Branched-chain fatty acids are found in some bacterial membrane lipids. Would their presence
increase or decrease the fluidity of the membranes (that is, give them a lower or higher melting
point)? Why?
Answer
(a) The number of cis double bonds (stearic acid, 18:0; oleic, 18:1; linoleic, 18:2; linolenic,
18:3). Each cis double bond causes a bend in the hydrocarbon chain, and bent chains are
less well packed than straight chains in a crystal lattice. The lower the extent of packing,
the lower the melting temperature.
(b) Six different triacylglycerols are possible: one with glycerol and only palmitic acid (PPP);
one with glycerol and only oleic acid (OOO); and four with glycerol and a mixture of
oleic and palmitic acids. Four mixed triacylglycerols are possible, because the three
carbons of glycerol are not equivalent: thus OOP and OPO are positional isomers, as are
POP and OPP. The greater the content of saturated fatty acid (P), the higher the melting
point. Thus, the order of melting points is OOO � OOP � OPO � POP � OPP � PPP.
See Table 10–1 and Figure 10–3 for information on how to draw the triacylglycerols.
(c) Branched-chain fatty acids will increase the fluidity of membranes (i.e., lower their
melting point) because they decrease the extent of packing possible within the
membrane. The effect of branches is similar to that of bends caused by double bonds.
3. Preparation of Béarnaise Sauce During the preparation of béarnaise sauce, egg yolks are incorpo-
rated into melted butter to stabilize the sauce and avoid separation. The stabilizing agent in the egg
yolks is lecithin (phosphatidylcholine). Suggest why this works.
Answer Lecithin (see Fig. 10–14 for structure), an amphipathic molecule, is an emulsifying agent,
solubilizing the fat (triacylglycerols) in butter. Lecithin is such a good emulsifying agent that it
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can be used to create a stable emulsion in a mixture that contains up to 75% oil. Mayonnaise,
too, is an emulsion created with egg yolks, with an oil:vinegar mixture in a 3:1 ratio.
4. Isoprene Units in Isoprenoids Geraniol, farnesol, and squalene are called isoprenoids, because they
are synthesized from five-carbon isoprene units. In each compound, circle the five-carbon units repre-
senting isoprene units (see Fig. 10–22).
Answer
5. Naming Lipid Stereoisomers The two compounds below are stereoisomers of carvone with quite
different properties; the one on the left smells like spearmint, and that on the right, like caraway.
Name the compounds using the RS system.
Answer Spearmint is (R)-carvone; caraway is (S)-carvone.
Geraniol
OH
Farnesol
OH
Squalene
Squalene
Spearmint
H2C
C
CH3 C
CH2
H
CH2
C
O C
CH3
CH
Caraway
H2C
C
H C
CH3
CH2
CH2
C
O C
CH3
CH
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6. RS Designations for Alanine and Lactate Draw (using wedge-bond notation) and label the (R)
and (S) isomers of 2-aminopropanoic acid (alanine) and 2-hydroxypropanoic acid (lactic acid).
Answer 
2-Aminopropanoic acid
(alanine)
E EE
C
H2N CH3
H
COOH
A
2-Hydroxypropanoic acid
(lactic acid)
E EE
C
HO CH3
H
COOH
A
(R)-2-Aminopropanoic acid
COOH
CH
CH3
�
NH3
(S)-2-Aminopropanoic acid
COOH
C
CH3
H
�
H3N
(R)-2-Hydroxypropanoic acid (S)-2-Hydroxypropanoic acid
COOH
C
H
OHH3C
OH
C
H
COOHH3C
7. Hydrophobic and Hydrophilic Components of Membrane Lipids A common structural feature of
membrane lipids is their amphipathic nature. For example, in phosphatidylcholine, the two fatty acid
chains are hydrophobic and the phosphocholine head group is hydrophilic. For each of the following
membrane lipids, name the components that serve as the hydrophobic and hydrophilic units: (a) phos-
phatidylethanolamine; (b) sphingomyelin; (c) galactosylcerebroside; (d) ganglioside; (e) cholesterol.
Answer
Hydrophobic unit(s) Hydrophilic unit(s)
(a) 2 Fatty acids Phosphoethanolamine
(b) 1 Fatty acid and the hydrocarbon Phosphocholine
chain of sphingosine
(c) 1 Fatty acid and the hydrocarbon D-Galactose
chain of sphingosine
(d) 1 Fatty acid and the hydrocarbon Several sugar molecules
chain of sphingosine
(e) Steroid nucleus and acyl side chain Alcohol group
8. Structure of Omega-6 Fatty Acid Draw the structure of the omega-6 fatty acid 16:1.
Answer 
9. Catalytic Hydrogenation of Vegetable Oils Catalytic hydrogenation, used in the food industry, con-
verts double bonds in the fatty acids of the oil triacylglycerols to OCH2OCH2O. How does this affect the
physical properties of the oils?
Answer It reduces double bonds, which increases the melting point of lipids containing the 
fatty acids.
10. Alkali Lability of Triacylglycerols A common procedure for cleaning the grease trap in a sink is to
add a product that contains sodium hydroxide. Explain why this works.
O
�O
C
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Answer Triacylglycerols, a component of grease (consisting largely of animal fats), are
hydrolyzed by NaOH to form glycerol and the sodium salts of free fatty acids, a process known as
saponification. The fatty acids form micelles, which are more water-soluble than triacylglycerols.
11. Deducing Lipid Structure from Composition Compositional analysis of a certain lipid shows
that it has exactly one mole of fatty acid per mole of inorganic phosphate. Could this be a glyc-
erophospholipid? A ganglioside? A sphingomyelin?
Answer It could only be a sphingolipid (sphingomyelin). Sphingomyelin has one fatty acid
molecule and a phosphocholine molecule attached to the sphingosine backbone, for a ratio 
of fatty acid to inorganic phosphate of 1:1. Glycerophospholipids have two fatty acyl chains
and a head group attached to a phosphoglycerol molecule. Unless the head group included 
additional phosphate groups, the ratio of fatty acid to inorganic phosphate would be 2:1.
(Phosphatidylinositol 4,5-bisphosphate would have a ratio of 2:3; cardiolipin, a ratio of 4:2.)
Gangliosides do not contain inorganic phosphate.
12. Deducing Lipid Structure from Molar Ratio of Components Complete hydrolysis of a glyc-
erophospholipid yields glycerol, two fatty acids (16:1(�9) and 16:0), phosphoric acid, and serine in the
molar ratio 1:1:1:1:1. Name this lipid and draw its structure.
Answer
13. Impermeability of Waxes What property of the waxy cuticles that cover plant leaves makes the cuti-
cles impermeable to water?
Answer Long, saturated acyl chains, nearly solid at air
temperature, form a hydrophobic layer
in which a polar compound such as H2O cannot dissolve or diffuse.
14. The Action of Phospholipases The venom of the Eastern diamondback rattler and the Indian cobra
contains phospholipase A2, which catalyzes the hydrolysis of fatty acids at the C-2 position of glyc-
erophospholipids. The phospholipid breakdown product of this reaction is lysolecithin (lecithin is phos-
phatidylcholine). At high concentrations, this and other lysophospholipids act as detergents, dissolving
the membranes of erythrocytes and lysing the cells. Extensive hemolysis may be life-threatening.
(a) All detergents are amphipathic. What are the hydrophilic and hydrophobic portions of
lysolecithin?
(b) The pain and inflammation caused by a snake bite can be treated with certain steroids. What is
the basis of this treatment?
(c) Though the high levels of phospholipase A2 can be deadly, this enzyme is necessary for a variety
of normal metabolic processes. What are these processes?
Answer
(a) The free OOH group on C-2 and the phosphocholine head group on C-3 are the
hydrophilic portions; the fatty acid on C-1 of the lysolecithin is the hydrophobic portion.
Chapter 10 Lipids S-115
Phosphatidylserine
O
CH2 CH2
CH
NH3C
COO�
H
CH2O C
O
C
O
O
O�
P
O
O
�
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(b) Certain steroids such as prednisone inhibit the action of phospholipase A2, the enzyme
that releases the fatty acid arachidonate from the C-2 position of some membrane
glycerophospholipids. Arachidonate is converted to a variety of eicosanoids, some of
which cause inflammation and pain.
(c) Phospholipase A2 is necessary to release arachidonate from certain membrane
glycerophopholipids. Arachidonate is a precursor of other eicosanoids that have vital
protective functions in the body. The enzyme is also important in digestion, breaking
down dietary glycerophospholipids.
15. Lipids in Blood Group Determination We note in Figure 10–15 that the structure of glycosphin-
golipids determines the blood groups A, B, and O in humans. It is also true that glycoproteins deter-
mine blood groups. How can both statements be true?
Answer The part of the membrane lipid that determines blood type is the oligosaccharide in
the head group of the membrane sphingolipids (see Fig. 10–15, p. 355). This same oligosaccha-
ride is attached to certain membrane glycoproteins, which also serve as points of recognition by
the antibodies that distinguish blood groups.
16. Intracellular Messengers from Phosphatidylinositols When the hormone vasopressin stimulates
cleavage of phosphatidylinositol 4,5-bisphosphate by hormone-sensitive phospholipase C, two products
are formed. What are they? Compare their properties and their solubilities in water, and predict
whether either would diffuse readily through the cytosol.
Answer Phosphatidylinositol 4,5-bisphosphate is a membrane lipid. The two products of
cleavage are a diacylglycerol and inositol 1,4,5-trisphosphate (IP3). Diacylglycerol is not
water-soluble and remains in the membrane, acting as a second messenger. The IP3 is highly
polar and very soluble in water; it readily diffuses in the cytosol, acting as a soluble second
messenger.
17. Storage of Fat-Soluble Vitamins In contrast to water-soluble vitamins, which must be part of our
daily diet, fat-soluble vitamins can be stored in the body in amounts sufficient for many months. Sug-
gest an explanation for this difference.
Answer Unlike water-soluble compounds, lipid-soluble compounds are not readily mobilized—
that is, they do not readily pass into aqueous solution. The body’s lipids provide a reservoir for
storage of lipid-soluble vitamins. Water-soluble vitamins cannot be stored and are rapidly re-
moved from the blood by the kidneys.
18. Hydrolysis of Lipids Name the products of mild hydrolysis with dilute NaOH of (a) 1-stearoyl-2,
3-dipalmitoylglycerol; (b) 1-palmitoyl-2-oleoylphosphatidylcholine.
Answer Mild hydrolysis cleaves the ester linkages between glycerol and fatty acids, forming 
(a) glycerol and the sodium salts of palmitic and stearic acids; (b) D-glycerol 3-phosphocholine
and the sodium salts of palmitic and oleic acids.
19. Effect of Polarity on Solubility Rank the following in order of increasing solubility in water: a 
triacylglycerol, a diacylglycerol, and a monoacylglycerol, all containing only palmitic acid.
Answer Solubilities: monoacylglycerol � diacylglycerol � triacylglycerol. Increasing the num-
ber of palmitic acid moieties increases the proportion of the molecule that is hydrophobic.
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20. Chromatographic Separation of Lipids A mixture of lipids is applied to a silica gel column, and
the column is then washed with increasingly polar solvents. The mixture consists of phosphatidylser-
ine, phosphatidylethanolamine, phosphatidylcholine, cholesteryl palmitate (a sterol ester), sphin-
gomyelin, palmitate, n-tetradecanol, triacylglycerol, and cholesterol. In what order will the lipids elute
from the column? Explain your reasoning.
Answer Because silica gel is polar, the most hydrophobic lipids elute first, the most hydrophilic
last. The neutral lipids elute first: cholesteryl palmitate and triacylglycerol. Cholesterol and 
n-tetradecanol, neutral but somewhat more polar, elute next. The neutral phospholipids
phosphatidylcholine and phosphatidylethanolamine follow. Sphingomyelin, neutral but slightly
more polar, elutes after the neutral phospholipids. The negatively charged phosphatidylserine
and palmitate elute last—phosphatidylserine first because it is larger and has a lower charge-to-
mass ratio.
21. Identification of Unknown Lipids Johann Thudichum, who practiced medicine in London about
100 years ago, also dabbled in lipid chemistry in his spare time. He isolated a variety of lipids from
neural tissue, and characterized and named many of them. His carefully sealed and labeled vials of
isolated lipids were rediscovered many years later.
(a) How would you confirm, using techniques not available to Thudichum, that the vials labeled
“sphingomyelin” and “cerebroside” actually contain these compounds?
(b) How would you distinguish sphingomyelin from phosphatidylcholine by chemical, physical, or 
enzymatic tests?
Answer
(a) First, create an acid hydrolysate of each compound. Sphingomyelin yields sphingosine, fatty
acids, phosphocholine, choline, and phosphate. Cerebroside yields sphingosine, fatty acids,
and sugars, but no phosphate. Subject each hydrolysate to chromatography (gas-liquid or
silica gel thin-layer chromatography) and compare the result with known standards.
(b) On strong alkaline hydrolysis, sphingomyelin yields sphingosine, fatty acids, and
phosphocholine; phosphatidylcholine yields glycerol, fatty acids, and phosphocholine. 
The distinguishing features are the presence of sphingosine in sphingomyelin and
glycerol in phosphatidylcholine, which can be detected on thin-layer chromatograms of
each hydrolysate compared against known standards. The hydrolysates could also be
distinguished by their reaction with the Sanger reagent (1-fluoro-2,4-dinitrobenzene,
FDNB); only the sphingosine in the sphingomyelin hydrolysate has a primary amine that
would react with FDNB to form a colored product. Alternatively, enzymatic treatment of
the two samples with phospholipase A1 or A2 would release free fatty acids from
phosphatidylcholine, but not from sphingomyelin.
22. Ninhydrin to Detect Lipids on TLC Plates Ninhydrin reacts specifically with primary amines to
form a purplish-blue product. A thin-layer chromatogram of rat liver phospholipids is sprayed with
ninhydrin, and the
color is allowed to develop. Which phospholipids can be detected in this way?
Answer Phosphatidylethanolamine and phosphatidylserine; they are the only phospholipids
that have primary amine groups that can react with ninhydrin.
Data Analysis Problem
23. Determining the Structure of the Abnormal Lipid in Tay-Sachs Disease Box 10–2, Figure 1,
shows the pathway of breakdown of gangliosides in healthy (normal) individuals and individuals with
certain genetic diseases. Some of the data on which the figure is based were presented in a paper by
Lars Svennerholm (1962). Note that the sugar Neu5Ac, N-acetylneuraminic acid, represented in the
Box 10–2 figure as a purple �, is a sialic acid.
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Svennerholm reported that “about 90% of the monosialiogangliosides isolated from normal human
brain” consisted of a compound with ceramide, hexose, N-acetylgalactosamine, and N-acetylneu-
raminic acid in the molar ratio 1:3:1:1.
(a) Which of the gangliosides (GM1 through GM3 and globoside) in Box 10–2, Figure 1, fits this de-
scription? Explain your reasoning.
(b) Svennerholm reported that 90% of the gangliosides from a patient with Tay-Sachs had a molar
ratio (of the same four components given above) of 1:2:1:1. Is this consistent with the Box 10–2
figure? Explain your reasoning.
To determine the structure in more detail, Svennerholm treated the gangliosides with neu-
raminidase to remove the N-acetylneuraminic acid. This resulted in an asialoganglioside that was much
easier to analyze. He hydrolyzed it with acid, collected the ceramide-containing products, and deter-
mined the molar ratio of the sugars in each product. He did this for both the normal and the Tay-Sachs
gangliosides. His results are shown below.
Ganglioside Ceramide Glucose Galactose Galactosamine
Normal
Fragment 1 1 1 0 0
Fragment 2 1 1 1 0
Fragment 3 1 1 1 1
Fragment 4 1 1 2 1
Tay-Sachs
Fragment 1 1 1 0 0
Fragment 2 1 1 1 0
Fragment 3 1 1 1 1
(c) Based on these data, what can you conclude about the structure of the normal ganglioside? Is
this consistent with the structure in Box 10–2? Explain your reasoning.
(d) What can you conclude about the structure of the Tay-Sachs ganglioside? Is this consistent with
the structure in Box 10–2? Explain your reasoning.
Svennerholm also reported the work of other researchers who “permethylated” the normal 
asialoganglioside. Permethylation is the same as exhaustive methylation: a methyl group is added to
every free hydroxyl group on a sugar. They found the following permethylated sugars: 2,3,6-trimethyl-
glycopyranose; 2,3,4,6-tetramethylgalactopyranose; 2,4,6-trimethylgalactopyranose; and 4,6-dimethyl-
2-deoxy-2-aminogalactopyranose.
(e) To which sugar of GM1 does each of the permethylated sugars correspond? Explain your
reasoning.
(f) Based on all the data presented so far, what pieces of information about normal ganglioside struc-
ture are missing?
Answer
(a) GM1 and globoside. Both glucose and galactose are hexoses, so “hexose” in the molar ra-
tio refers to glucose + galactose. The ratios for the four gangliosides are: GM1, 1:3:1:1;
GM2, 1:2:1:1; GM3, 1:2:0:1; globoside, 1:3:1:0. 
(b) Yes. The ratio matches GM2, the ganglioside expected to build up in Tay-Sachs disease
(see Box 10–2, Fig. 1). 
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(c) This analysis is similar to that used by Sanger to determine the amino acid sequence of
insulin. The analysis of each fragment reveals only its composition, not its sequence,
but because each fragment is formed by sequential removal of one sugar, we can draw
conclusions about sequence. The structure of the normal asialoganglioside is ceramide–
glucose–galactose–galactosamine–galactose, consistent with Box 10–2 (excluding
Neu5Ac, removed before hydrolysis). 
(d) The Tay-Sachs asialoganglioside is ceramide–glucose–galactose–galactosamine, consis-
tent with Box 10–2. 
(e) The structure of the normal asialoganglioside, GM1, is: ceramide–glucose (2 —OH
involved in glycosidic links; 1 —OH involved in ring structure; 3 —OH (2,3,6) free for
methylation)– galactose (2 —OH in links; 1 —OH in ring; 3 —OH (2,4,6) free for
methylation)–galactosamine (2 —OH in links; 1 —OH in ring; 1 —NH2 instead of 
an —OH; 2 —OH (4,6) free for methylation)–galactose (1 —OH in link; 1 —OH in ring;
4 —OH (2,3,4,6) free for methylation).
(f) Two key pieces of information are missing: What are the linkages between the sugars?
Where is Neu5Ac attached?
Reference
Svennerholm, L. (1962) The chemical structure of normal human brain and Tay-Sachs gangliosides. Biochem. Biophys. Res. Comm.
9, 436–441.
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