Ch. 10, 11 Biochem short answer

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1

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?

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.

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.

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?

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

Catalytic Hydrogenation of Vegetable Oils Catalytic hydrogenation, used in the food industry, converts
double bonds in the fatty acids of the oil triacylglycerols to OCH2OCH2O. How does this affect the
physical properties of the oils?

It reduces double bonds, which increases the melting point of lipids containing the
fatty acids.

4

Rank the following in order of increasing solubility in water: a
triacylglycerol, a diacylglycerol, and a monoacylglycerol, all containing only palmitic acid.

Solubilities: monoacylglycerol  diacylglycerol  triacylglycerol. Increasing the number
of palmitic acid moieties increases the proportion of the molecule that is hydrophobic.

5

Lipid bilayers formed between two aqueous phases have
this important property: they form two-dimensional sheets, the edges of which close upon each other
and undergo self-sealing to form liposomes.

(a) What properties of lipids are responsible for this property of bilayers? Explain.

(b) What are the consequences of this property for the structure of biological membranes?

(a) Lipids that form bilayers are amphipathic molecules: they contain hydrophilic and
hydrophobic regions. In order to minimize the hydrophobic area exposed to the water
surface, these lipids form two-dimensional sheets, with the hydrophilic regions exposed
to water and the hydrophobic regions buried in the interior of the sheet. Furthermore, to
avoid exposing the hydrophobic edges of the sheet to water, lipid bilayers close upon
themselves. Similarly, if the sheet is perforated, the hole will seal because the membrane
is semifluid.

(b) These sheets form the closed membrane surfaces that envelop cells and compartments
within cells (organelles).

6

The carbon–carbon bond distance for single-bonded carbons
such as those in a saturated fatty acyl chain is about 1.5 Å. Estimate the length of a single molecule of
palmitate in its fully extended form. If two molecules of palmitate were placed end to end, how would
their total length compare with the thickness of the lipid bilayer in a biological membrane?

Given that the C–C bond length is 0.15 nm and that the bond angle of tetrahedral
carbon is 109, the distance between the first and third carbons in an acyl chain (calculated by
trigonometry) is about 0.24 nm. For palmitate (16:0), the length of the extended chain is
about 8  0.24 nm  2 nm. Two palmitate chains end to end (as in a bilayer) would extend 4 nm.
This is about the thickness of a lipid bilayer.

7

Location of a Membrane Protein The following observations are made on an unknown membrane
protein, X. It can be extracted from disrupted erythrocyte membranes into a concentrated salt solution,
and it can be cleaved into fragments by proteolytic enzymes. Treatment of erythrocytes with proteolytic
enzymes followed by disruption and extraction of membrane components yields intact X. However, treatment
of erythrocyte “ghosts” (which consist of just plasma membranes, produced by disrupting the cells
and washing out the hemoglobin) with proteolytic enzymes followed by disruption and extraction yields
extensively fragmented X. What do these observations indicate about the location of X in the plasma
membrane? Do the properties of X resemble those of an integral or peripheral membrane protein?

Because protein X can be removed by salt treatment, it must be a peripheral
membrane protein. Inability to digest the protein with proteases unless the membrane has
been disrupted indicates that protein X is located internally, bound to the inner surface of the
erythrocyte plasma membrane.

8

Membrane lipids in tissue samples obtained from different parts of the
leg of a reindeer have different fatty acid compositions. Membrane lipids from tissue near the hooves
contain a larger proportion of unsaturated fatty acids than those from tissue in the upper leg. What is
the significance of this observation?

The temperature of body tissues at the extremities, such as near the hooves, is
generally lower than that of tissues closer to the center of the body. To maintain fluidity, as
required by the fluid-mosaic model, membranes at lower temperatures must contain a higher
percentage of polyunsaturated fatty acids: a higher content of unsaturated fatty acids lowers
the melting point of lipid mixtures.

9

At pH 7, tryptophan crosses a lipid bilayer at about one-thousandth the
rate of indole, a closely related compound:

Suggest an explanation for this observation.

At pH 7, tryptophan exists as a zwitterion (having a positive and negative charge),
whereas indole is uncharged. The movement of the less polar indole through the hydrophobic
core of the bilayer is more energetically favorable.

10

A helical wheel is a two-dimensional representation of a helix, a
view along its central axis (see Fig. 11–29b; see also Fig. 4–4d). Use the helical wheel diagram below
to determine the distribution of amino acid residues in a helical segment with the sequence –Val–
Asp–Arg–Val–Phe–Ser–Asn–Val–Cys–Thr–His–Leu–Lys–Thr–Leu–Gln–Asp–Lys–

What can you say about the surface properties of this helix? How would you expect the helix to be
oriented in the tertiary structure of an integral membrane protein?

A helical wheel is a two-dimensional representation of a helix obtained by projecting
the helix down its central axis. An  helix contains 3.6 residues per turn, so each amino acid
in the helix lies 100 around the axis from the previous residue: (360/turn)/(3.6 residues/turn) 
100 per residue. For the 18 amino acid helix considered here, the 18 vertices are separated by
20 increments. If there were a 19th residue, it would lie under the first residue on the projection,
but five turns down the helix: 5 turns  0.54 nm/turn (pitch for an  helix)  2.70 nm “behind”
residue 1. To complete the diagram, follow the lines from residue 1 to residue 2, and so on,
numbering the residues. Then, using the sequence given, label each residue with its one-letter
abbreviation and a characterization of its R group properties—P for polar, and N for nonpolar.

The “top” side of the helix contains only hydrophobic side chains, while the other surfaces are
polar or charged; this is an amphipathic helix. As an integral membrane protein, it is likely to
dip its hydrophobic surface into the lipid bilayer but expose the other surfaces to the aqueous
phase. An alternative arrangement might be to cluster, say, 10 helices, one from each of 10 subunits,
around a central hydrophilic core, while exposing only the hydrophobic surface to the
lipid bilayer.