Plasma Membrane
Help Questions
AP Biology › Plasma Membrane
Two cell membranes contain identical transport proteins but differ in phospholipid tail composition. Membrane Q has a higher proportion of saturated fatty acids; Membrane R has a higher proportion of unsaturated fatty acids. At the same temperature, lateral movement of membrane proteins is slower in Membrane Q than in Membrane R. Which membrane feature best explains the reduced protein mobility in Membrane Q?
Which feature best explains the slower lateral movement of proteins in Membrane Q?
More saturated tails create larger gaps, allowing proteins to move faster through the bilayer
More saturated tails pack tightly, decreasing membrane fluidity and limiting protein diffusion
More saturated tails shorten phospholipids, forcing proteins to unfold and slow transport
More saturated tails increase membrane permeability, causing proteins to leak out of the bilayer
More saturated tails increase negative charge, forming ionic bonds that immobilize proteins
Explanation
This question assesses the skill of analyzing plasma membrane structure and transport. The reduced protein mobility in Membrane Q results from its higher proportion of saturated fatty acids, which pack tightly and decrease overall membrane fluidity. This gel-like state limits lateral diffusion of embedded proteins compared to Membrane R's more fluid, unsaturated tails that allow easier movement. Both membranes have identical proteins, so tail saturation directly affects the bilayer's physical properties at the same temperature. A tempting distractor is choice D, which mistakenly claims saturated tails create larger gaps for faster movement, reversing the actual effect of reduced fluidity on protein diffusion. When comparing protein mobility, examine how fatty acid saturation influences membrane fluidity and lateral dynamics.
A phospholipid bilayer contains embedded transport proteins. The hydrophobic tails create a barrier to charged particles, while some proteins form selective pores lined with polar amino acids. A student compares movement of K+ across membranes with and without a functional K+ channel protein. The concentration of K+ is higher outside than inside. Which feature best explains why K+ moves into the cell only when the channel is functional?
A K+ channel provides a hydrophilic pore that reduces interaction with the bilayer core
K+ enters only when the channel makes the membrane thicker, increasing permeability
K+ enters only when the membrane becomes more saturated, increasing fluidity
K+ crosses by simple diffusion because charged ions pass through hydrophobic tails easily
Phospholipid heads bind K+ and flip to the inner leaflet to deliver the ion
Explanation
This question assesses the skill of analyzing plasma membrane structure and transport. The functional K+ channel protein forms a hydrophilic pore lined with polar amino acids, allowing charged K+ ions to pass through without interacting with the hydrophobic tails of the bilayer. Without the channel, K+ cannot cross the hydrophobic barrier easily, even with a favorable concentration gradient from outside to inside. This selective pore reduces the energy barrier for ion movement, explaining entry only when the channel is present. A tempting distractor is choice A, which suggests K+ crosses by simple diffusion through tails, stemming from the misconception that charged ions can dissolve in nonpolar environments. When analyzing ion movement, distinguish between simple diffusion and facilitated transport by considering the charge and the role of specific proteins.
A plasma membrane is a phospholipid bilayer with embedded proteins. Some proteins act as carriers that bind a specific solute and change shape to move it across the membrane. In a test, a polar amino acid is transported into cells faster when its carrier protein is present, even though the amino acid concentration is equal on both sides of the membrane. The bilayer’s hydrophobic interior remains unchanged. Which feature best explains the increased amino acid movement when the carrier is present?
Amino acids diffuse rapidly through the hydrophobic core because they are polar
Phospholipid heads rotate inward, exposing water to the core and moving amino acids
Carrier proteins create a hydrophilic route by binding the solute and changing conformation
Embedded proteins convert polar solutes into nonpolar solutes so they cross the bilayer
Carrier proteins increase membrane thickness, which increases permeability to polar solutes
Explanation
This question assesses the skill of analyzing plasma membrane structure and transport. Carrier proteins bind the polar amino acid and change conformation to provide a hydrophilic route across the hydrophobic interior, facilitating movement even without a concentration gradient. This mechanism increases transport rate when the carrier is present, as the bilayer's core otherwise repels polar solutes. The unchanged hydrophobic interior emphasizes the carrier's role in overcoming this barrier. A tempting distractor is choice B, which incorrectly states amino acids diffuse rapidly through the core due to polarity, based on the misconception that polar molecules can cross nonpolar regions without assistance. For facilitated transport questions, examine how protein conformational changes enable solute movement independent of gradients.
Two membranes have the same phospholipid bilayer composition but differ in cholesterol content. Membrane A has high cholesterol; Membrane B has low cholesterol. At a low temperature, small nonpolar molecules diffuse more readily through Membrane A than through Membrane B. Cholesterol is embedded among phospholipid tails. Which membrane property change best explains the observation?
Which change would most likely explain higher permeability in the high-cholesterol membrane at low temperature?
Cholesterol binds to nonpolar molecules and carries them across by active transport
Cholesterol reduces tight packing of phospholipid tails at low temperature, maintaining fluidity
Cholesterol increases the polarity of the bilayer interior, dissolving nonpolar molecules faster
Cholesterol increases phospholipid head charge, pulling nonpolar molecules across the surface
Cholesterol forms open channels that allow nonpolar molecules to pass through the membrane
Explanation
This question assesses the skill of analyzing plasma membrane structure and transport. At low temperatures, Membrane A with high cholesterol allows better diffusion of small nonpolar molecules by reducing tight packing of phospholipid tails, thus maintaining membrane fluidity. Cholesterol embeds among the tails and prevents crystallization-like solidification that would occur in low-cholesterol Membrane B, increasing permeability. Both membranes have the same phospholipids, so cholesterol's moderating effect on fluidity explains the difference. A tempting distractor is choice C, which falsely claims cholesterol forms open channels, confusing its structural role with that of transport proteins. To assess permeability at varying temperatures, analyze how cholesterol influences bilayer fluidity and packing.
A cell membrane is a phospholipid bilayer with a hydrophobic interior and embedded proteins. Researchers compare transport of two solutes: oxygen (small and nonpolar) and sucrose (large and polar). The membrane contains no sucrose-specific transport proteins. Both solutes are present at higher concentration outside the cell than inside. Which feature best explains why oxygen enters the cell more readily than sucrose?
Oxygen is nonpolar and can dissolve in the hydrophobic core, allowing simple diffusion
Sucrose is polar and therefore diffuses rapidly through the hydrophobic tails
Sucrose crosses faster because large molecules move more quickly through membranes
Oxygen enters faster because phospholipid heads bind oxygen and pull it inward
Oxygen requires carrier proteins because it is nonpolar and cannot cross the bilayer
Explanation
This question assesses the skill of analyzing plasma membrane structure and transport. Oxygen, being small and nonpolar, can readily dissolve in the hydrophobic interior of the phospholipid bilayer and cross by simple diffusion down its concentration gradient. Sucrose, however, is large and polar, making it unable to pass easily through the hydrophobic core without specific transport proteins, which are absent in this membrane. This difference in polarity and the lack of sucrose transporters explain why oxygen enters more readily despite both having favorable gradients. A tempting distractor is choice B, which claims sucrose diffuses rapidly due to its polarity, based on the misconception that polar molecules interact favorably with hydrophobic regions. To solve transport comparison questions, classify solutes by size and polarity to predict their diffusion mechanisms across bilayers.
An experiment measures water movement across two cell membranes. Both membranes are phospholipid bilayers with hydrophilic heads and hydrophobic tails; however, Membrane X contains many aquaporin proteins, and Membrane Y contains few. Water is polar and crosses the hydrophobic interior slowly by simple diffusion. The cells are transferred to a hypertonic solution, creating a water concentration gradient. Which change would most likely increase the rate of water loss from cells with Membrane Y?
Increase aquaporin abundance to provide more hydrophilic channels for water
Replace phospholipids with proteins so the membrane becomes fully hydrophilic
Decrease the number of aquaporins so fewer water molecules can cross
Increase phospholipid head polarity to prevent water from entering the bilayer
Increase the number of carbohydrate chains on the outer surface of the membrane
Explanation
This question assesses the skill of analyzing plasma membrane structure and transport. Increasing aquaporin abundance in Membrane Y would provide more hydrophilic channels, allowing polar water molecules to cross the hydrophobic interior more rapidly via facilitated diffusion. Membrane X already has many aquaporins, enabling faster water movement, while Membrane Y's few aquaporins limit the rate, especially under a hypertonic gradient driving water loss. This change directly addresses the barrier posed by the hydrophobic tails, which slow simple diffusion of water. A tempting distractor is choice A, which suggests decreasing aquaporins, based on the misconception that fewer channels would somehow enhance water flow rather than restrict it. When tackling osmosis-related questions, focus on how protein channels modulate the permeability of polar molecules across lipid bilayers.
Two cells have plasma membranes with identical phospholipid composition and thickness. Cell X has many transmembrane carrier proteins specific for glucose; Cell Y has far fewer of these carriers. Both cells are placed in a solution with glucose concentration higher outside than inside. Glucose enters Cell X faster than Cell Y, even though glucose is polar. Which membrane feature best explains the higher glucose uptake rate?
Which feature best explains why glucose enters Cell X faster than Cell Y?
Glucose crosses faster in Cell X because polar molecules diffuse best through lipid cores
Carrier proteins in Cell X reduce extracellular glucose concentration, increasing diffusion force
More carrier proteins increase phospholipid tail saturation, raising glucose solubility
A higher density of specific carrier proteins increases facilitated diffusion of glucose
Fewer carriers in Cell Y increase membrane fluidity, preventing glucose from entering
Explanation
This question assesses the skill of analyzing plasma membrane structure and transport. Glucose enters Cell X faster because it has a higher density of specific carrier proteins that facilitate diffusion of the polar glucose molecules down their concentration gradient. Despite identical phospholipid composition, the abundance of carriers in Cell X provides more pathways for glucose, which cannot easily cross the hydrophobic bilayer without assistance. This explains the higher uptake rate even though glucose is polar and the gradient is the same for both cells. A tempting distractor is choice B, which mistakenly states more carriers increase tail saturation to raise glucose solubility, ignoring that saturation affects fluidity, not polar solute solubility in lipids. To compare transport rates, consider how the number of specific transport proteins influences facilitated diffusion of polar molecules.
A student constructs two liposomes from identical phospholipids. Liposome X has no membrane proteins; Liposome Y includes a transmembrane protein that forms a continuous water-filled pore. When placed in a solution containing a small charged ion, the ion enters Liposome Y rapidly but shows minimal entry into Liposome X. The phospholipid bilayer interior is hydrophobic. Which feature best explains the difference in ion permeability between the two liposomes?
The pore provides a hydrophilic route that bypasses the hydrophobic core, increasing ion entry
The bilayer interior is hydrophilic, so ions enter X but not Y due to protein crowding
The pore decreases membrane thickness, making the bilayer tails shorter and more charged
The pore blocks diffusion by binding ions permanently, preventing movement into the vesicle
The pore converts ions into nonpolar molecules, allowing them to dissolve in the tails
Explanation
This question tests understanding of plasma membrane structure and transport through protein channels. The correct answer is B because the transmembrane protein creates a water-filled pore that provides a continuous hydrophilic pathway across the membrane, allowing charged ions to move through without encountering the hydrophobic bilayer interior that would otherwise exclude them. The comparison between identical liposomes with and without the pore protein clearly demonstrates that ions require a hydrophilic route to cross efficiently. Answer A is incorrect because it claims the bilayer interior is hydrophilic, which represents a fundamental misconception - the phospholipid tails create a hydrophobic core that repels charged particles. When analyzing ion permeability, remember that charged particles need hydrophilic pathways to cross hydrophobic barriers.
The plasma membrane is composed of a phospholipid bilayer with hydrophobic tails that form a nonpolar interior. Some phospholipids have unsaturated fatty acid tails that introduce kinks, reducing tight packing. A researcher compares two artificial membranes: Membrane X has mostly saturated tails, and Membrane Y has more unsaturated tails. Both lack transport proteins. When exposed to the same small nonpolar solute, Membrane Y shows faster diffusion of the solute.
Which feature best explains the higher permeability of Membrane Y?
Unsaturated tails increase the membrane’s overall charge, attracting nonpolar solutes into the bilayer
Saturated tails bind nonpolar solutes and shuttle them across without proteins
Unsaturated tails convert the bilayer interior into a hydrophilic region that excludes nonpolar solutes
Unsaturated tails pack less tightly, increasing bilayer fluidity and diffusion of nonpolar solutes
Saturated tails create hydrophilic pores that specifically block nonpolar solutes
Explanation
This question requires analyzing how plasma membrane lipid composition affects transport properties. The correct answer is A because unsaturated fatty acid tails have kinks that prevent tight packing of phospholipids, creating a more fluid bilayer with more space between molecules for nonpolar solutes to move through. The experiment shows that Membrane Y, with more unsaturated tails, allows faster diffusion of nonpolar solutes compared to Membrane X with mostly saturated (straight) tails that pack tightly together. This increased fluidity and spacing in the hydrophobic core makes it easier for nonpolar molecules to dissolve into and diffuse through the membrane. Answer D is incorrect because unsaturated tails don't convert the bilayer interior into a hydrophilic region—the interior remains hydrophobic regardless of saturation, and this hydrophobic nature is what allows nonpolar solutes to cross. When comparing membrane permeability, remember that lipid saturation affects packing and fluidity: unsaturated = more fluid = faster diffusion for molecules that cross through the lipid bilayer.
A researcher builds liposomes from phospholipid bilayers containing embedded proteins. One set of liposomes includes many aquaporin proteins; the other set has the same phospholipids but no aquaporins. When placed in a hypotonic solution, liposomes with aquaporins swell more rapidly. The phospholipid heads are polar and the interior fatty acid tails are nonpolar in both sets. Which membrane component difference best accounts for the faster swelling?
Which feature best explains the increased water movement into the aquaporin-containing liposomes?
Aquaporins provide hydrophilic channels that allow rapid water passage across the bilayer
Aquaporins decrease the number of phospholipids, shortening the diffusion distance for water
Aquaporins increase membrane surface charge, pulling water across by electrostatic attraction
Aquaporins make the phospholipid tails more polar, increasing water solubility in the core
Aquaporins convert water into ions, which diffuse through the hydrophobic interior
Explanation
This question assesses the skill of analyzing plasma membrane structure and transport. The faster swelling of aquaporin-containing liposomes in a hypotonic solution results from aquaporins forming hydrophilic channels that enable rapid water passage through the otherwise hydrophobic bilayer core. Without aquaporins, water crosses slowly due to the nonpolar fatty acid tails repelling polar water molecules, but aquaporins provide a polar pathway that bypasses this barrier. Both sets have identical phospholipids, so the difference is solely due to these embedded proteins facilitating osmosis. A tempting distractor is choice B, which wrongly claims aquaporins make tails more polar to increase water solubility, confusing the role of proteins with altering lipid properties. When studying osmosis across membranes, identify how specific proteins like aquaporins enhance transport of polar molecules like water.