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AP Biology Help: Membrane Permeability

Review real example questions for Membrane Permeability in AP Biology.

Question 1 / 10

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A phospholipid bilayer has a hydrophobic interior. A solute's permeability depends on how well it can enter this nonpolar region. Small, nonpolar molecules cross readily; polar molecules cross slowly; large polar molecules cross very slowly; ions cross least. Compare two uncharged molecules: ribose (a 5-carbon sugar with multiple hydroxyl groups) and isopropanol (a 3-carbon alcohol with one hydroxyl group). No transport proteins are present.

Which molecule would most likely be more permeable across the bilayer?

All questions

Question 1

A phospholipid bilayer has a hydrophobic interior. A solute's permeability depends on how well it can enter this nonpolar region. Small, nonpolar molecules cross readily; polar molecules cross slowly; large polar molecules cross very slowly; ions cross least. Compare two uncharged molecules: ribose (a 5-carbon sugar with multiple hydroxyl groups) and isopropanol (a 3-carbon alcohol with one hydroxyl group). No transport proteins are present.

Which molecule would most likely be more permeable across the bilayer?

  1. Ribose, because it has more oxygen atoms to interact with the membrane
  2. Isopropanol, because it is less polar and smaller than ribose (correct answer)
  3. Ribose, because sugars are used by cells and therefore diffuse easily
  4. Isopropanol, because polar molecules cross faster than nonpolar molecules
  5. Both, because neither molecule is charged

Explanation: This question assesses the skill of analyzing membrane permeability based on molecular properties in a phospholipid bilayer. Isopropanol is more permeable than ribose because it is smaller and less polar with only one hydroxyl group, allowing better solubility in the hydrophobic interior, while ribose has multiple hydroxyls making it highly polar and larger. Without transport proteins, simple diffusion favors less polar molecules. Both are uncharged, but isopropanol's properties reduce the energy barrier more effectively. A tempting distractor is ribose because sugars are used by cells (choice C), but this reflects the misconception that biological relevance affects physical diffusion, whereas permeability depends on molecular traits. To analyze similar problems, evaluate size and polarity together, as smaller, less polar molecules diffuse faster across bilayers.

Question 2

A model membrane is composed of a phospholipid bilayer with a hydrophobic interior. Molecules that are small and nonpolar tend to partition into the lipid core and diffuse across, whereas polar molecules interact strongly with water and are less soluble in the membrane interior. Charged molecules are surrounded by hydration shells and experience a large energetic barrier to entering the hydrophobic region. Consider two uncharged molecules of similar size: ethanol (contains a hydroxyl group) and propane (a hydrocarbon). No channels or carriers are present.

Which explanation best accounts for propane crossing the membrane more readily than ethanol?

  1. Propane is nonpolar, so it dissolves in the hydrophobic core more easily (correct answer)
  2. Propane is larger, so it is pushed through by collisions more often
  3. Ethanol is polar, so it must use ATP to cross any membrane
  4. Ethanol is uncharged, so it is excluded by the membrane surface
  5. Propane crosses faster because water repels it into the membrane

Explanation: This question assesses the skill of analyzing membrane permeability based on molecular properties in a phospholipid bilayer. Propane crosses more readily than ethanol because it is nonpolar, allowing it to partition easily into the hydrophobic interior, while ethanol's hydroxyl group makes it polar and less soluble in the lipid core. The similar size of the molecules highlights that polarity is the key differentiator, as nonpolar molecules dissolve better without interacting strongly with water. No channels or carriers mean simple diffusion depends on solubility in the membrane, favoring propane. A tempting distractor is that ethanol is polar so must use ATP (choice C), but this is wrong due to the misconception that all polar crossings require energy, whereas simple diffusion is passive but slower for polar molecules. To analyze similar problems, compare polarity first for molecules of similar size, as nonpolar ones have higher permeability in bilayers.

Question 3

D-glucose uptake saturates without ATP; L-glucose uptake is linear with concentration. This shows that

  1. D-glucose uses a carrier (correct answer)
  2. L-glucose uses a carrier
  3. Both use the same carrier
  4. Both use simple diffusion

Explanation: Saturable uptake means D-glucose is limited by carrier proteins that can be occupied, while linear uptake means L-glucose enters only by simple diffusion without a carrier. The tempting wrong answer is that both use simple diffusion, but that would make D-glucose uptake linear, not saturable.

Question 4

Compared with a similar-size ion, O2 crosses the lipid bilayer faster because O2 can

  1. Pass through an ion channel
  2. Dissolve in the lipid core (correct answer)
  3. Use the sodium-potassium pump
  4. Move through aquaporin pores

Explanation: Oxygen is small and nonpolar, so it dissolves readily in the hydrophobic lipid core of the membrane and diffuses across. A similar-size ion is charged and cannot enter that hydrophobic core easily, which is why ion channels are tempting but actually are needed for ions, not for O2.

Question 5

Red cells lyse in 0.3 M urea but not in 0.15 M NaCl. Why?

  1. Urea crosses; water follows (correct answer)
  2. NaCl crosses; water leaves
  3. Urea cannot cross the membrane
  4. Urea is actively pumped inward

Explanation: Urea readily crosses the red cell membrane, so it enters the cell and raises internal solute concentration; water follows by osmosis, causing swelling and lysis. NaCl is impermeable, so 0.15 M NaCl does not cause net water entry. The tempting mistake is thinking urea cannot cross the membrane, but it is permeable, which is exactly why water follows.

Question 6

At 40°C, cholesterol lowers membrane water permeability. The most likely mechanism is that cholesterol

  1. Pumps water out of cells
  2. Opens gated ion channels
  3. Compacts hydrophobic tails (correct answer)
  4. Keeps lipid tails more fluid

Explanation: Cholesterol fills gaps between phospholipid tails and compacts them, so water has fewer routes to sneak through the membrane. At 40°C membranes are already fluid; cholesterol stiffens and tightens the hydrophobic core. The tempting mistake is thinking cholesterol always makes membranes more fluid, but it only does that at cooler temperatures; here it reduces permeability by compaction.

Question 7

To keep net diffusion rate unchanged when the concentration gradient doubles, the membrane area must be

  1. Quadrupled
  2. Doubled
  3. Unchanged
  4. Halved (correct answer)

Explanation: Diffusion rate depends on the product of membrane area and concentration gradient. If the gradient doubles, halving the area keeps that product unchanged. The tempting error is to double the area, but that would make the rate four times larger, not unchanged.

Question 8

A cell membrane is modeled as a phospholipid bilayer with no transport proteins. Two solutes are compared for passive movement across the membrane: solute X is a 6-carbon sugar with multiple hydroxyl (–OH) groups and no net charge; solute Y is a 4-carbon hydrocarbon with no polar groups and no charge. Both are present at the same concentration outside the cell. The bilayer core is hydrophobic, so nonpolar molecules have higher solubility in it than polar molecules. Polar groups form favorable interactions with water, which reduces their tendency to enter the nonpolar interior. Differences in permeability can be inferred from polarity and size alone under these conditions.

  1. Solute X, because being uncharged is sufficient for rapid diffusion through the hydrophobic core.
  2. Solute Y, because nonpolar molecules dissolve in the bilayer core more readily than polar molecules. (correct answer)
  3. Solute X, because multiple –OH groups make it more compatible with phospholipid tails.
  4. Solute Y, because smaller molecules always cross faster regardless of polarity differences.
  5. Both cross at similar rates, because equal external concentration eliminates permeability differences.

Explanation: This question assesses the skill of analyzing membrane permeability based on molecular properties in a phospholipid bilayer without transport proteins. The correct answer is solute Y because it is a nonpolar hydrocarbon, which dissolves readily in the hydrophobic bilayer core, as noted in the stimulus where nonpolar molecules have higher solubility than polar ones. Solute X, a 6-carbon sugar with multiple –OH groups, is polar and forms favorable interactions with water, reducing its tendency to enter the nonpolar interior despite being uncharged. Although solute Y is smaller, its nonpolarity is the key factor enhancing permeability over the larger, polar solute X under equal concentration conditions. A tempting distractor is choice A, which wrongly claims that being uncharged is sufficient for rapid diffusion, embodying the misconception that lack of charge overrides polarity effects in hydrophobic environments. A transferable strategy is to prioritize nonpolarity over size when comparing uncharged molecules' ability to cross lipid bilayers by passive diffusion.

Question 9

A phospholipid bilayer without proteins separates two chambers. Equal concentrations of glyceraldehyde (90 Da, polar uncharged) and O2_2 (32 Da, nonpolar) are placed on one side. Which statement best explains which solute accumulates on the opposite side first?

  1. Glyceraldehyde arrives first because polar molecules interact with phospholipid heads
  2. O2_2 arrives first because nonpolar molecules cross the hydrophobic core readily (correct answer)
  3. Glyceraldehyde arrives first because it is larger and moves down gradients faster
  4. Both arrive equally because diffusion depends only on concentration difference
  5. Neither arrives because uncharged molecules cannot cross a bilayer

Explanation: This question tests the skill of analyzing membrane permeability based on solute properties in a phospholipid bilayer. O₂ arrives first because it is nonpolar, crossing the hydrophobic core readily down its concentration gradient. Glyceraldehyde is polar uncharged, facing resistance that slows its accumulation on the opposite side. The stimulus describes equal starting concentrations and no proteins, focusing on diffusion rates. A tempting distractor is choice C, suggesting larger size speeds diffusion, but this reflects the misconception that mass increases gradient-driven movement. For transferable strategy, always predict nonpolar solutes accumulate fastest in diffusion setups, considering polarity next for timing outcomes.

Question 10

In an experiment, a pure phospholipid bilayer is exposed to equal concentrations of K+^+ (39 Da, charged) and argon gas (40 Da, nonpolar). Their masses are similar. Which molecule would most likely cross the membrane faster by simple diffusion?

  1. K+^+, because it is slightly smaller and therefore diffuses more rapidly
  2. Argon, because nonpolar molecules pass readily through the hydrophobic core (correct answer)
  3. K+^+, because charged particles are attracted to phospholipid tails
  4. Both equally, because they have nearly the same molecular mass
  5. Neither, because diffusion across membranes requires ATP hydrolysis

Explanation: This question tests the skill of analyzing membrane permeability based on solute properties in a phospholipid bilayer. Argon crosses faster because it is nonpolar, allowing easy passage through the hydrophobic core despite similar mass to K⁺. K⁺ is charged, making it highly impermeable as ions are repelled by the nonpolar interior. The stimulus notes similar masses and no proteins, emphasizing that polarity determines rate over size for diffusion. A tempting distractor is choice A, suggesting K⁺ is faster due to slight size difference, but this reflects the misconception that size overrides charge barriers in bilayers. For transferable strategy, always prioritize nonpolarity and lack of charge for rapid diffusion, using mass as a tiebreaker only for similar properties.