AP Biology Quiz: Membrane Permeability

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Membrane PermeabilityQuestion 1 of 20

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?

Ribose, because it has more oxygen atoms to interact with the membrane

Isopropanol, because it is less polar and smaller than ribose

Ribose, because sugars are used by cells and therefore diffuse easily

Isopropanol, because polar molecules cross faster than nonpolar molecules

Both, because neither molecule is charged

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AP Biology Quiz

AP Biology Quiz: Membrane Permeability

Practice Membrane Permeability in AP Biology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Membrane Permeability, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Biology.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

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

Ribose, because it has more oxygen atoms to interact with the membrane
Isopropanol, because it is less polar and smaller than ribose (correct answer)
Ribose, because sugars are used by cells and therefore diffuse easily
Isopropanol, because polar molecules cross faster than nonpolar molecules
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?

Propane is nonpolar, so it dissolves in the hydrophobic core more easily (correct answer)
Propane is larger, so it is pushed through by collisions more often
Ethanol is polar, so it must use ATP to cross any membrane
Ethanol is uncharged, so it is excluded by the membrane surface
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

D-glucose uses a carrier (correct answer)
L-glucose uses a carrier
Both use the same carrier
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

Pass through an ion channel
Dissolve in the lipid core (correct answer)
Use the sodium-potassium pump
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?

Urea crosses; water follows (correct answer)
NaCl crosses; water leaves
Urea cannot cross the membrane
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

Pumps water out of cells
Opens gated ion channels
Compacts hydrophobic tails (correct answer)
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

Quadrupled
Doubled
Unchanged
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.

Solute X, because being uncharged is sufficient for rapid diffusion through the hydrophobic core.
Solute Y, because nonpolar molecules dissolve in the bilayer core more readily than polar molecules. (correct answer)
Solute X, because multiple –OH groups make it more compatible with phospholipid tails.
Solute Y, because smaller molecules always cross faster regardless of polarity differences.
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 O $_2$ (32 Da, nonpolar) are placed on one side. Which statement best explains which solute accumulates on the opposite side first?

Glyceraldehyde arrives first because polar molecules interact with phospholipid heads
O $_2$ arrives first because nonpolar molecules cross the hydrophobic core readily (correct answer)
Glyceraldehyde arrives first because it is larger and moves down gradients faster
Both arrive equally because diffusion depends only on concentration difference
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?

K $^+$ , because it is slightly smaller and therefore diffuses more rapidly
Argon, because nonpolar molecules pass readily through the hydrophobic core (correct answer)
K $^+$ , because charged particles are attracted to phospholipid tails
Both equally, because they have nearly the same molecular mass
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.

Question 11

A synthetic vesicle is made only of phospholipids, creating a hydrophobic membrane core. Molecules that are small and nonpolar cross more readily than molecules that are large, polar, or charged. Consider glycerol (small but polar due to three hydroxyl groups) and methane (very small and nonpolar). Neither molecule carries a net charge. No transport proteins are present, and temperature is constant.

Which molecule would most likely have the higher permeability across the vesicle membrane?

Glycerol, because its hydroxyl groups interact with the membrane surface
Methane, because it is small and nonpolar (correct answer)
Glycerol, because it is smaller than most sugars
Methane, because nonpolar molecules cannot dissolve in water
Both, because neither is charged

Explanation: This question assesses the skill of analyzing membrane permeability based on molecular properties in a phospholipid bilayer. Methane has higher permeability than glycerol because it is very small and nonpolar, allowing it to cross the hydrophobic core readily, while glycerol's three hydroxyl groups make it polar and less soluble in lipids. The vesicle's pure phospholipid composition and lack of transport proteins mean simple diffusion favors nonpolar molecules. Both are uncharged, but methane's nonpolarity overcomes glycerol's polarity despite similar small size. A tempting distractor is glycerol because it is smaller than most sugars (choice C), but this stems from the misconception that size is the only factor, ignoring how polarity hinders membrane solubility. To analyze similar problems, rank molecules by nonpolarity and small size for permeability in pure bilayers, as these properties facilitate diffusion.

Question 12

A phospholipid bilayer with no proteins is tested with three solutes: NH $_3$ (17 Da, uncharged, polar), NH $_4^+$ (18 Da, charged), and N $_2$ (28 Da, nonpolar). All are present at equal concentration. Which solute would most likely cross the bilayer at the highest rate?

NH $_4^+$ (18 Da, charged)
NH $_3$ (17 Da, uncharged, polar)
N $_2$ (28 Da, nonpolar) (correct answer)
NH $_4^+$ , because it is slightly heavier and diffuses more forcefully
All three, because all are under 30 Da

Explanation: This question tests the skill of analyzing membrane permeability based on solute properties in a phospholipid bilayer. N₂ crosses at the highest rate because it is nonpolar, enabling rapid diffusion through the hydrophobic core. NH₃ is polar uncharged and NH₄⁺ is charged, both facing barriers to entry that N₂ avoids. The stimulus lists small sizes under 30 Da and no proteins, underscoring nonpolarity's advantage. A tempting distractor is choice A (NH₄⁺), due to slight heaviness, but this ignores the misconception that mass trumps charge in permeability. For transferable strategy, always rank nonpolar gases highest, followed by polar uncharged, with charged solutes lowest in bilayer diffusion.

Question 13

A phospholipid bilayer is impermeable to most ions because the hydrophobic interior disfavors charged species. Two nitrogen-containing solutes are compared: nitrous oxide (N2O), which is small and relatively nonpolar, and ammonium (NH4+), which is charged. Both are present at equal concentration. No transport proteins are present.

Which solute would most likely cross the bilayer more readily by simple diffusion?

Ammonium (NH4+), because it is small and contains hydrogen
Nitrous oxide (N2O), because it is uncharged and relatively nonpolar (correct answer)
Ammonium (NH4+), because positive charge attracts it to the membrane
Both, because nitrogen-containing molecules cross membranes readily
Neither, because diffusion requires a carrier for any solute

Explanation: This question assesses the skill of analyzing membrane permeability based on molecular properties in a phospholipid bilayer. Nitrous oxide (N2O) crosses more readily than ammonium (NH4+) because it is uncharged and relatively nonpolar, dissolving in the hydrophobic interior, while NH4+ is charged and impermeable without proteins. Equal concentrations highlight charge as the barrier in simple diffusion. The bilayer's impermeability to ions disfavors NH4+. A tempting distractor is NH4+ because it is small and contains hydrogen (choice A), but this reflects the misconception that size and composition override charge, whereas charge is prohibitive. To analyze similar problems, always select uncharged, nonpolar molecules over ions for faster bilayer diffusion.

Question 14

A phospholipid bilayer separates extracellular fluid from cytosol. The membrane interior is hydrophobic, so diffusion across it favors small, nonpolar molecules. Polar molecules cross slowly, and charged molecules cross extremely slowly because charge is energetically unfavorable in the hydrophobic core. Compare alanine in its zwitterionic form (has both positive and negative charges at physiological pH) and alanine methyl ester (neutral, less polar). No transport proteins are present.

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

Alanine (zwitterion), because it is small and can hydrogen-bond
Alanine methyl ester, because it is neutral and less polar (correct answer)
Alanine (zwitterion), because charges are attracted to lipid tails
Alanine methyl ester, because esters are always charged in water
Both, because they are derived from the same amino acid

Explanation: This question assesses the skill of analyzing membrane permeability based on molecular properties in a phospholipid bilayer. Alanine methyl ester is more permeable than alanine zwitterion because it is neutral and less polar, allowing better dissolution in the hydrophobic core, while the zwitterion's charges create a high energetic barrier. At physiological pH, the zwitterion's positive and negative charges disfavor entry into nonpolar regions. No transport proteins mean simple diffusion strongly prefers uncharged forms. A tempting distractor is alanine zwitterion because charges attract to lipid tails (choice C), but this reflects the misconception that charges aid solubility, whereas they prevent it in hydrophobic environments. To analyze similar problems, evaluate charged versus neutral forms, as neutral molecules cross bilayers more readily.

Question 15

A phospholipid bilayer membrane is tested with different solutes. The membrane interior is hydrophobic, so permeability increases as solutes become smaller and less polar. Ions are especially impermeable because their charge is stabilized by water and unfavorable in the membrane core. Compare chloride ion (Cl−) and chlorine gas (Cl2). Both contain chlorine atoms, but one is charged and the other is nonpolar.

Which solute would most likely diffuse across the bilayer more rapidly?

Chloride ion (Cl−), because it is smaller than Cl2
Chlorine gas (Cl2), because it is nonpolar and uncharged (correct answer)
Chloride ion (Cl−), because ions dissolve in water and thus enter membranes
Both, because they are made of the same element
Neither, because gases cannot cross a liquid membrane

Explanation: This question assesses the skill of analyzing membrane permeability based on molecular properties in a phospholipid bilayer. Chlorine gas (Cl2) diffuses more rapidly than chloride ion (Cl−) because it is nonpolar and uncharged, dissolving easily in the hydrophobic interior, while Cl− is charged and repelled by the nonpolar core. Permeability increases with decreasing polarity and charge, making Cl2 favored despite both containing chlorine. No proteins mean ions are highly impermeable. A tempting distractor is Cl− because it is smaller (choice A), but this ignores the misconception that size overrides charge, whereas charge is a major barrier. To analyze similar problems, distinguish charged from uncharged forms, as uncharged versions cross bilayers much faster.

Question 16

A phospholipid bilayer is exposed to two solutes at equal concentration. The membrane interior is hydrophobic, so permeability depends on how well a solute can enter that nonpolar region. Compare two molecules of similar size: acetic acid in its uncharged form (CH3COOH) and acetate (CH3COO−), which carries a negative charge. Assume the pH conditions keep one solute mostly uncharged and the other charged. No proteins are present.

Which solute would most likely cross the bilayer faster by simple diffusion?

Acetate (CH3COO−), because charge increases solubility in the membrane
Acetic acid (CH3COOH), because it is uncharged and less hydrophilic (correct answer)
Acetate (CH3COO−), because it is smaller than acetic acid
Both, because they differ only by one proton
Neither, because acids cannot diffuse through lipid bilayers

Explanation: This question assesses the skill of analyzing membrane permeability based on molecular properties in a phospholipid bilayer. Acetic acid (CH3COOH) crosses faster than acetate (CH3COO−) because it is uncharged and less hydrophilic, allowing easier entry into the hydrophobic core, while the charged acetate is stabilized by water and repelled. pH conditions maintain the charge difference, emphasizing charge's role in permeability. Equal concentrations and no proteins focus on simple diffusion. A tempting distractor is acetate because charge increases solubility (choice A), but this reflects the misconception that charge helps in nonpolar environments, whereas it hinders it. To analyze similar problems, compare protonated and deprotonated forms, favoring the uncharged for higher bilayer permeability.

Question 17

A phospholipid bilayer (no proteins) separates two solutions containing equal concentrations of solute A (methane, 16 Da, nonpolar) and solute B (formaldehyde, 30 Da, polar uncharged). Which molecule would most likely have the higher permeability across the bilayer?

Formaldehyde, because polar molecules dissolve in the aqueous head groups
Methane, because small nonpolar molecules partition into the lipid interior (correct answer)
Formaldehyde, because it is larger and therefore crosses more effectively
Both equally, because both are uncharged and under 50 Da
Neither, because gases cannot diffuse through liquid membranes

Explanation: This question tests the skill of analyzing membrane permeability based on solute properties in a phospholipid bilayer. Methane has higher permeability because it is small and nonpolar, partitioning easily into the lipid interior for fast diffusion. Formaldehyde is polar uncharged, which reduces its solubility in the hydrophobic core despite being small. The stimulus specifies sizes under 50 Da and no proteins, highlighting nonpolarity's key role in permeability. A tempting distractor is choice A, claiming formaldehyde is faster due to polarity, but this ignores the misconception that polar molecules dissolve well in nonpolar lipids. For transferable strategy, always assess nonpolarity first for bilayer permeability, as it facilitates entry more than size or weak polarity.

Question 18

A membrane made only of phospholipids is exposed to equal concentrations of ribose (150 Da, polar uncharged) and benzene (78 Da, nonpolar). No proteins are present. Which molecule would most likely be more permeable through the membrane?

Ribose, because it is larger and therefore more likely to enter the bilayer
Benzene, because nonpolar molecules dissolve in the hydrophobic interior (correct answer)
Ribose, because uncharged polar molecules cross faster than nonpolar molecules
Both equally, because neither carries a net charge
Neither, because aromatic rings prevent diffusion through membranes

Explanation: This question tests the skill of analyzing membrane permeability based on solute properties in a phospholipid bilayer. Benzene is more permeable because it is nonpolar, dissolving easily in the hydrophobic interior despite ribose being uncharged. Ribose's polarity and larger size hinder its diffusion through the lipid core. The stimulus provides sizes and polarities with no proteins, emphasizing nonpolarity over size. A tempting distractor is choice A, favoring ribose's size, but this reflects the misconception that larger polar molecules cross better than smaller nonpolar ones. For transferable strategy, always prioritize nonpolar solutes for high permeability, assessing size only after polarity in pure bilayers.

Question 19

A lipid bilayer membrane lacks transport proteins. The hydrophobic core favors diffusion of small, nonpolar molecules. Polar molecules can cross only slowly because they are poorly soluble in the membrane interior, and ions are effectively excluded due to charge and hydration shells. Consider ammonia (NH3), which is small and polar, and neon (Ne), which is small and nonpolar. Both are uncharged in this scenario.

Which molecule would most likely diffuse across the membrane faster?

Ammonia (NH3), because it is polar and attracted to phospholipid heads
Neon (Ne), because it is small and nonpolar (correct answer)
Ammonia (NH3), because small molecules always cross quickly
Neon (Ne), because it is a gas and gases require no diffusion
Both diffuse equally because they have similar diameters

Explanation: This question assesses the skill of analyzing membrane permeability based on molecular properties in a phospholipid bilayer. Neon (Ne) diffuses across the membrane faster than ammonia (NH3) because it is small and nonpolar, dissolving readily in the hydrophobic core, while ammonia is polar and less soluble despite its small size. The lack of transport proteins means simple diffusion depends on lipid compatibility, favoring nonpolar atoms like neon. Both are uncharged, but neon's inert nonpolar nature gives it an edge over ammonia's polarity. A tempting distractor is ammonia because small molecules always cross quickly (choice C), but this ignores the misconception that size overrides polarity, whereas polarity slows diffusion in hydrophobic environments. To analyze similar problems, compare nonpolarity alongside size for diffusion rates in bilayers, as nonpolar molecules cross faster.

Question 20

A protein-free phospholipid bilayer is tested with two molecules of similar size: methanol (32 Da, polar uncharged) and oxygen (32 Da, nonpolar). Which molecule would most likely have greater permeability through the bilayer?

Methanol, because polarity increases solubility in lipid tails
Oxygen, because nonpolar molecules enter the hydrophobic interior more readily (correct answer)
Methanol, because equal mass means polarity does not affect diffusion
Both equally, because both are small molecules
Neither, because alcohols and gases require channels to cross membranes

Explanation: This question tests the skill of analyzing membrane permeability based on solute properties in a phospholipid bilayer. Oxygen has greater permeability because it is nonpolar, entering the hydrophobic interior more readily than the polar methanol. Both are small with equal mass, but polarity hinders methanol's diffusion. The stimulus emphasizes similar sizes and no proteins, isolating polarity's effect. A tempting distractor is choice A, claiming polarity increases solubility, but this ignores the misconception that polar molecules favor nonpolar lipids. For transferable strategy, always compare polarity directly when sizes match, favoring nonpolar for faster bilayer crossing.