Download CHEM* 356 F2000 Review Assignment #7 ANSWERS Dr. F.J.

CHEM* 356 F2000
Review Assignment #7 ANSWERS
Dr. F.J. Sharom
Cell membrane structure; glycoproteins, glycolipids, surface antigens
1.
What chemical composition would you expect for a typical plasma membrane?
A high lipid
B high protein C no carbohydrate
D no nucleic acid
X
2.
Cerebrosides are glycolipids which contain:
A choline
3.
B ceramide
X
C phosphate
D sphingomyelin
E O-acyl groups
N-acetylglucosamine residues found in glycoproteins are often linked to the side chains of:
A serine
B alanine
C aspartate
D asparagine
X
E threonine
4.
Lipid bilayers have an important property: they form 2-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?
Answer:
The hydrophobic tails of phospholipids are exposed to water at the edges of a bilayer sheet. It is
thermodynamically unfavorable to have the hydrophobic acyl chains exposed to water. The system
will, therefore, spontaneously move in the direction of minimizing exposed edges. The easiest way
to do this is for the edges to close upon themselves to form a vesicle.
b)
What are the consequences of this property with respect to the structure of cell membranes?
Answer:
Cells, and cellular organelles, all of which are bounded by membranes, form closed structures, rather
than sheets with exposed edges
5.
A common structural feature of membrane lipid molecules is their amphipathic nature. For
each of the following membrane lipids, name the components that serve as the hydrophobic and
hydrophilic units:
A phosphatidylethanolamine
B cholesterol
D galactosylcerebroside
E a ganglioside
C sphingomyelin
Answer:
hydrophilic
hydrophobic
A PE
phosphoethanolamine
2 fatty acids
B cholesterol
-OH group
hydrocarbon ring skeleton
C sphingomyelin
phosphocholine
ceramide
(sphingosine + 1 fatty acid)
D galactocerebroside
D-galactose
ceramide
E a ganglioside
oligosaccharide
(several sugars)
ceramide
6.
Why can't triacylglycerols be significant components of lipid bilayers?
Answer:
Lipids that form bilayers are amphipathic; they have a polar headgroup and nonpolar acyl chains.
Triacylglycerols lack a polar headgroup, so they cannot orient themselves to form a bilayer (To form
a bilayer, the non-polar acyl chains of triglycerides would have to be buried in the interior, and the
glycerol moiety would have to be at the surface in contact with the water.)
7.
The venom of the Indian cobra contains the enzyme phospholipase C, which catalyses the
hydrolysis of the fatty acid at the C-2 position of glycerophospholipids. The phospholipid
breakdown product produced is known as lysophosphatidylcholine (lyso-PC).
a)
What are the hydrophilic and hydrophobic portions of lyso-PC?
Answer:
Hydrophilic:
Hydrophobic:
b)
phosphocholine and -OH group at C2
one acyl chain at C1
Can you predict the properties of lyso-PC from its structure?
Answer:
Lyso-PC should be a detergent-like molecule, and will form micelles in solution, rather than bilayers
like PC.
c)
What effects do you think lyso-PC would have on biological membranes, and how might this
relate to the toxicity of cobra venom?
Answer:
Lyso-PC acts like a detergent, and will break up and dissolve cell membranes. This would clearly
be very damaging to cells, and would result in their lysis. In fact, the action of phospholipase C in
cobra venom causes extensive hemolysis of red blood cells, which can be life-threatening.
8.
Explain why transverse diffusion (flip-flop) is 109 times slower than lateral diffusion in a lipid
Answer:
For a phospholipid to flip-flop from one side of the membrane to the other means that the charged
polar headgroup must be moved through the hydrophobic interior of the bilayer. Clearly, this is
highly unfavourable thermodynamically, so it takes place at a very low rate. Lateral diffusion, on the
other hand, simply involves two adjacent phospholipids "changing places", and it occurs at a high
rate.
9.
You have made the following observations on a protein, named novellin, recently discovered
in the membrane of lymphocytes. Novellin can be extracted from disrupted lymphocyte
membranes with a concentrated salt solution, and following purification, the protein can then
be cleaved into fragments using the protease enzyme trypsin. If intact lymphocytes are treated
with trypsin, followed by disruption and salt extraction of the membrane, novellin is recovered
intact. However, if cells are disrupted first, and then treated with trypsin, the recovered
novellin is extensively fragmented. What do these observations tell you about the type of
membrane protein novellin is, and where it is located?
Answer:
Novellin can be removed from the membrane by high salt treatment, which suggests that it is a
peripheral protein, rather than an integral protein, which would require detergent treatment for
extraction.
Purified novellin can be cleaved by trypsin, so the protein clearly has some sites where cleavage can
take place (Lys and Arg residues). Since novellin is not cleaved by trypsin in intact cells, this implies
that it must be located on the cytosolic side of the membrane, where trypsin cannot reach it (trypsin
is a large protein, and cannot cross the cell membrane). This is confirmed by the observation that if
cells are disrupted first, to allow trypsin to gain access to the inside of the cell, then novellin is indeed
cleaved. Thus, we can say that novellin is probably a peripheral protein located on the cytosolic side
of the membrane.
10.
The lipid portion of a typical bilayer is about 3 nm thick.
a)
Calculate the number of residues of an α-helix that will just span this distance (rise per residue
= 0.15 nm). How many turns of the α-helix is this?
Answer:
rise per residue = distance travelled along helix for one residue = 0.15 nm
residues needed to span 3 nm membrane = 3.0 nm/0.15 nm = 20 residues
an α-helix has 3.6 residues/turn (check your text-book)
! # of turns needed to span membrane =20 ÷ 3.6 = 5.6 turns
b)
The human epidermal growth factor receptor has a single transmembrane helix. Can you find
it in the following sequence:
···· RGPKIPSIATGMVGALLLLVVALGIGILFMRRRH ····
Answer:
The sequence RRR is highly +vely charged, and probably defines the C-terminal end of the
transmembrane helix. Moving 19 residues up the sequence gives us:
MVGALLLLVVALGIGILFM
This seems like a reasonable candidate for the membrane-spanning sequence. However, we cannot
be sure, since the upstream sequence IPSIATG is also somewhat hydrophobic, and only then do we
find charged residues (RGPK), which clearly cannot be part of the membrane-spanning sequence.
11.
The receptor for the hormone epinephrine in animal cells has a molecular mass of 64 kDa, and
is known to have seven transmembrane segments.
a)
Show that a protein this size is capable of spanning the membrane seven times.
Answer:
The "average" MW of an amino acid is 100 Da. Thus a protein of MW 64 kDa will contain ~640
amino acid residues.
We calculated in question 10 above that only ~20 amino acids are needed to span the membrane. For
7 membrane-spanning segments, we would need a minimum of ~7 × 20 = 140 residues. So the
receptor is more than large enough to contain 7 membrane-spanning segments.
b)
Given the amino acid sequence of the protein, what strategy could you use to locate regions
that might form the seven transmembrane helices?
Answer:
As we saw in question 10 above, we can find potential membrane-spanning segments by searching
for stretches of hydrophobic amino acids around 20 residues long. This could be done by hand,
although it would be a tedious process. In fact, specialized software programs exist that search the
amino acid sequences of large proteins to identify hydrophobic transmembrane sequences.
12.
Most hormones, such as peptide hormones, do not enter the cell, and exert their effects by
binding to cell surface receptor proteins. However, steroid hormones do so by binding to an
intracellular protein receptor. Indicate how differences in structure and properties of the two
types of hormones can explain this difference in mechanism.
Answer:
Peptide hormones are hydrophilic, and will not be able to diffuse across the plasma membrane.
Instead, they bind to a membrane protein receptor, and it is this receptor that produces a "signal" of
some sort on the cytosolic side of the cell.
Steroid hormones resemble cholesterol structurally; they have the same ring structure as their nucleus
(they share the same biosynthetic pathway). They are therefore quite hydrophobic. They can dissolve
in membranes and diffuse across them to the cytosolic side, where they interact with an intracellular
receptor. Thus, steroid hormones do not need a membrane protein receptor.
13.
Rh-positive fetal red blood cells can leak through into the mother's circulation during delivery,
triggering antibody production in the serum of Rh-negative mothers. In subsequent
pregnancies, these antibodies cross the placenta and lead to life-threatening agglutination of
red blood cells in the baby. Why are such mothers given an injection of anti-Rh antibodies
(Rhogam) immediately after each birth?
A to make antibodies to the anti-Rh antibodies
B to agglutinate the mother's red blood cells
C to agglutinate red blood cells from the baby
X
D to change the mother's blood type