Problems
http://edugen.wiley.com/edugen/courses/crs1900/rc/voet9301c09/dm9ld...
1. Does trans-oleic acid have a higher or lower melting point than
cis-oleic
acid? Explain.
Answer:
-Oleic trans-Oleic
acid has h as a higher melting point point because, becau se, in the solid solid state, its hydrocarbon chains pack together more tightly tightly than those of cis-oleic acid. triacylglycerolss could incorporate the t he fatty fatt y acids shown in Fig. 9-1? 2. How many different types of triacylglycerol Answer:
Of the 4 × 4 = 16 pairs of fatty acid residues at C1 and C3, only 10 are unique because a molecule with different substituents at C1 and C3 is identical to the molecule with with the reverse substitution order. However, C2 may have any a ny of the four substituents for a total of 4 × 10 = 40 different triacylglycerol triacylglycerols. s. 3. Which triacylglycerol yields more energy on oxidation: one containing three residues of linolenic acid or three residues of stearic acid? a cid? Answer:
The triacylglycerol containing the stearic acid residues yields more energy since it is fully reduced. glycerophospholipid that has a saturated saturat ed C 16 fatty acyl group at position 1, a 4. Draw the structure of a glycerophospholipid monounsaturated C18 fatty acyl group at position 2, and an ethanolami ethan olamine ne head group. Answer:
a re obtained obt ained when 1-palmitoyl-2-ol 1-palmitoyl-2-oleoyl-3-phosphati eoyl-3-phosphatidylseri dylserine ne is hydrolyzed hydrolyzed by (a) 5. What products are phospholipase phospholipase A 1; (b) phospholipase A2; (c) phospholipase phospholipase C; (d) phospholipase D? Answer: (a) Palmitic acid and 2-oleoyl-3-phosphatidylserine; (b) oleic acid and 1-palmitoyl-3-phosphatidylserine; (c) phosphoserine and 1-palmitoyl-2-oleoyl-glycerol; (d) serine and 1-palmitoyl-2-oleoyl-phosphatidic acid. 6. Which of the glycerophospholipid head groups listed in Table 9-2 can form hydrogen bonds? Answer:
All except choline can form hydrogen bonds.
Problems
http://edugen.wiley.com/edugen/courses/crs1900/rc/voet9301c09/dm9ld...
7. Does the phosphatidylglycerol “head group” of cardiolipin (Table 9-2) project out of a lipid bilayer like
other glycerophospholi glycerophospholipid pid head groups? Answer:
No; the two t wo acyl chains of the “ head group” are buried in the bilayer interior, interior, leaving a head group of diphosphoglycerol. 8. In some autoimmune diseases, an individual develops antibodies that recognize cell constituents such as DNA and phospholipids. Some of the antibodies react with both DNA and phospholipids. What is the structural basis for this cross-reactivity? Answer:
Both DNA and phospholipids phospholipids have exposed phosphate groups that are recognized by the antibodies. th eir effects by binding to cell-surface cell-surface receptors. 9. Most hormones, such as peptide hormones, exert their However, steroid hormones do so by binding to cytosolic receptors. How is this possible? Answer:
Steroid Steroid hormones, which are hydrophobic, can diffuse through the cell membrane membrane to reach their receptors. recept ors. Animals cannot synthesize linoleic linoleic acid (a precursor of arachidonic acid) and therefore th erefore must obtain this t his 10. Animals cells can survive in the absence of essential fatty acid from their diet. Explain why cultured animal cells linoleic acid. Answer:
Eicosanoids synthesized synthesized from arachidonic acid are necessary for intercellular intercellular communication. Cultured cells do not need such communication communication and therefore t herefore do not require linoleic linoleic acid. 11. Why can't triacylglycerols be significant components of lipid bilayers? Answer:
Triacylglycerols lack polar head groups, so they do not orient themselves in a bilayer with their acyl chains inward and their glycerol moiety moiety toward the surface. 12. Why would a bilayer containing only gangliosides be unstable? Answer:
The large oligosaccharide head groups of gangliosides would prevent efficient packing of the lipids in a bilayer. 13. When bacteria growing at 20°C are warmed to 30°C, are they more likely to synthesize membrane lipids with (a) saturated or unsaturated un saturated fatty acids, and (b) short-chain or long-chain fatty acids? Explain. Explain. Answer:
(a) Saturated; (b) long-chain. long-chain. By increasing the proportion of saturated and an d long-chain fatty acids, which have higher melting melting points, the bacteria ba cteria can maintain maintain constant membrane fluidity fluidity at the higher temperature. 14. (a) How many turns of an α helix are required to span a lipid bilayer (~30 Å across)? (b) What is the minimum number of residues required? (c) Why do most transmembrane helices contain more than the minimum number of residues?
Problems
http://edugen.wiley.com/edugen/courses/crs1900/rc/voet9301c09/dm9ld...
Answer: (a) (1 turn/5.4 Å)(30 Å) = 5.6 turns (b) (3.6 residues/turn)(5.6 turns) = 20 residues (c) The additional residues form a helix, which partially satisfies backbone hydrogen bonding
requirements, requirements, where the lipid head groups do not offer hydrogen bonding partners. 15. The distance between the C α atoms in a β sheet is ~3.5 Å. Can a single 9-residue segment with a β
conformation serve as the transmembrane transmembrane portion of an integral membrane membrane protein? prote in? Answer:
No. Although the β strand could span the bilayer, bilayer, a single single strand would be unstable because its backbone could not form the hydrogen bonds it would form with water in aqueous solution. solution. 16. Are the following lipid samples likely to correspond to the inner or outer leaflet of a eukaryotic plasma membrane? (a) 20% phosphatidylcholine, phosphatidylcholine, 15% 15 % phosphatidylserine, phosphatidylserine, 65% other lipids. lipids. (b) 35% phosphatidylcholine, 15% gangliosides, 5% cholesterol, 45% other lipids. Answer:
(a) Inner; (b) outer. ou ter. See Fig. Fig. 9-32. 17. Describe the labeling pattern of glycophorin A when a membrane-impermeable protein-labeling reagent is added to (a) a preparation of solubili solubilized zed erythrocyte proteins; (b) intact erythrocyte ghosts; and (c) erythrocyte ghosts that are initially initially leaky and then th en immediately immediately sealed and transferred t ransferred to a solution that does not contain cont ain the labeling reagent. reagent. Answer:
(a) Both the intra- and a nd extracell extrac ellular ular portions will will be labeled. (b) Only the extracell extrace llular ular portion will be labeled. (c) Only the intracellular portion will be labeled. signal peptidase peptidase that narrows its specificity specificity so that it cleaves only 18. Predict the effect of a mutation in signal between two Leu residues. Answer:
The mutant signal peptidase would cleave many preproteins within their signal peptides, which often contain Leu-Leu sequences. This would would not affect translocation into the ER, E R, since signal signal peptidase acts after the signal signal peptide enters the ER lumen. Proteins lacking the Leu-Leu sequence would retain their signal peptides. These proteins, and those with abnormally cleaved signal sequences, would be more likely likely to fold abnormally abnormally and therefore function abnormally. 19. Explain why a drug that interferes with the disassembly of a SNARE complex would block neurotransmission. Answer:
In order for a neuron to repeatedly release neurotransmitters, the components of its exocytotic machinery must be recycled. Following the fusion of synaptic vesicles with the plasma membrane, the four-helix SNARE complex is disassembled so that the Q-SNAREs remain in the plasma membrane while portions of the membrane containing R-SNAREs can be used to re-form synaptic vesicles. This recycling recycling process would not be possible possible if the R- and Q-SNAREs remained remained associated, and the neuron ne uron
Problems
http://edugen.wiley.com/edugen/courses/crs1900/rc/voet9301c09/dm9ld...
would eventually be unable to t o release neurotransmitters.
Copyright © 2009 John Wiley & Sons, Inc. All rights reserved.
