AP Biology Quiz: Membrane Permeability

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.

Question 1

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

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

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 4

1. Glyceraldehyde arrives first because polar molecules interact with phospholipid heads
2. O$_2$ 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 5

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.

Question 6

1. Glycerol, because its hydroxyl groups interact with the membrane surface
2. Methane, because it is small and nonpolar (correct answer)
3. Glycerol, because it is smaller than most sugars
4. Methane, because nonpolar molecules cannot dissolve in water
5. 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 7

1. NH$_4^+$ (18 Da, charged)
2. NH$_3$ (17 Da, uncharged, polar)
3. N$_2$ (28 Da, nonpolar) (correct answer)
4. NH$_4^+$, because it is slightly heavier and diffuses more forcefully
5. 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 8

1. Ammonium (NH4+), because it is small and contains hydrogen
2. Nitrous oxide (N2O), because it is uncharged and relatively nonpolar (correct answer)
3. Ammonium (NH4+), because positive charge attracts it to the membrane
4. Both, because nitrogen-containing molecules cross membranes readily
5. 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 9

1. Alanine (zwitterion), because it is small and can hydrogen-bond
2. Alanine methyl ester, because it is neutral and less polar (correct answer)
3. Alanine (zwitterion), because charges are attracted to lipid tails
4. Alanine methyl ester, because esters are always charged in water
5. 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 10

1. Chloride ion (Cl−), because it is smaller than Cl2
2. Chlorine gas (Cl2), because it is nonpolar and uncharged (correct answer)
3. Chloride ion (Cl−), because ions dissolve in water and thus enter membranes
4. Both, because they are made of the same element
5. 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 11

1. Acetate (CH3COO−), because charge increases solubility in the membrane
2. Acetic acid (CH3COOH), because it is uncharged and less hydrophilic (correct answer)
3. Acetate (CH3COO−), because it is smaller than acetic acid
4. Both, because they differ only by one proton
5. 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 12

1. Formaldehyde, because polar molecules dissolve in the aqueous head groups
2. Methane, because small nonpolar molecules partition into the lipid interior (correct answer)
3. Formaldehyde, because it is larger and therefore crosses more effectively
4. Both equally, because both are uncharged and under 50 Da
5. 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 13

1. Ribose, because it is larger and therefore more likely to enter the bilayer
2. Benzene, because nonpolar molecules dissolve in the hydrophobic interior (correct answer)
3. Ribose, because uncharged polar molecules cross faster than nonpolar molecules
4. Both equally, because neither carries a net charge
5. 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 14

1. Ammonia (NH3), because it is polar and attracted to phospholipid heads
2. Neon (Ne), because it is small and nonpolar (correct answer)
3. Ammonia (NH3), because small molecules always cross quickly
4. Neon (Ne), because it is a gas and gases require no diffusion
5. 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 15

1. Methanol, because polarity increases solubility in lipid tails
2. Oxygen, because nonpolar molecules enter the hydrophobic interior more readily (correct answer)
3. Methanol, because equal mass means polarity does not affect diffusion
4. Both equally, because both are small molecules
5. 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.

Question 16

1. Solute Y, because smaller size always increases permeability more than polarity
2. Solute X, because nonpolarity favors entry into the hydrophobic interior (correct answer)
3. Solute Y, because uncharged molecules cross regardless of polarity
4. Solute X, because larger molecules diffuse faster due to higher kinetic energy
5. Neither, because only ions can pass through phospholipid bilayers

Explanation: This question tests the skill of analyzing membrane permeability based on solute properties in a phospholipid bilayer. Solute X diffuses more rapidly because its nonpolarity allows it to dissolve readily in the hydrophobic interior, overcoming its larger size compared to the polar solute Y. Although solute Y is smaller, its strong polarity hinders entry into the nonpolar core, reducing its permeability. The stimulus details no transport proteins and uncharged nature, highlighting that nonpolar solutes often permeate faster even if larger. A tempting distractor is choice A, favoring smaller size always, but this reflects the misconception that size dominates over nonpolarity in lipid solubility. For transferable strategy, always evaluate nonpolarity as the primary factor for bilayer diffusion, with size secondary for comparing similar polarities.

Question 17

1. Sucrose, because it is uncharged
2. Diethyl ether, because it is relatively nonpolar (correct answer)
3. Sucrose, because large molecules have higher momentum
4. Diethyl ether, because polar molecules are excluded from water
5. Both, because neither is an ion

Explanation: This question assesses the skill of analyzing membrane permeability based on molecular properties in a phospholipid bilayer. Diethyl ether has greater permeability than sucrose because it is relatively nonpolar, dissolving well in the hydrophobic interior, while sucrose is large and highly polar with many hydroxyl groups, making it poorly permeable. Both are uncharged, but ether's nonpolarity overcomes sucrose's size and polarity issues. No transport proteins mean simple diffusion disfavors large polar molecules. A tempting distractor is sucrose because it is uncharged (choice A), but this ignores the misconception that lack of charge alone ensures permeability, whereas size and polarity also matter greatly. To analyze similar problems, prioritize nonpolar over large polar uncharged molecules for bilayer permeability assessments.

Question 18

1. Formamide, because it is polar and can interact with water
2. Carbon monoxide (CO), because it is small and relatively nonpolar (correct answer)
3. Formamide, because it contains nitrogen which increases diffusion
4. Carbon monoxide (CO), because it has a dipole and is therefore excluded
5. Both, because uncharged molecules cross at the same rate

Explanation: This question assesses the skill of analyzing membrane permeability based on molecular properties in a phospholipid bilayer. Carbon monoxide (CO) has greater permeability than formamide because it is small and relatively nonpolar, allowing easy passage through the hydrophobic interior, while formamide is strongly polar due to its carbonyl and amine groups. Both are uncharged, but CO's lower polarity enhances solubility in lipids. Constant temperature and gradients ensure the comparison focuses on molecular traits without proteins. A tempting distractor is formamide because it is polar and interacts with water (choice A), but this reflects the misconception that water interaction aids membrane crossing, whereas it actually hinders lipid solubility. To analyze similar problems, prioritize small, less polar molecules for higher permeability in pure bilayers.

Question 19

1. Urea, because it is small
2. CO2, because it is small and nonpolar (correct answer)
3. Urea, because it can hydrogen-bond with phospholipid tails
4. CO2, because it has two oxygen atoms and is therefore polar
5. Both diffuse equally because they have similar molar mass

Explanation: This question assesses the skill of analyzing membrane permeability based on molecular properties in a phospholipid bilayer. CO2 diffuses across the bilayer faster than urea because it is small and nonpolar, enabling easy dissolution in the hydrophobic interior, while urea is polar due to electronegative atoms and interacts strongly with water. The absence of membrane proteins ensures permeability relies on simple diffusion, where nonpolarity trumps polarity even for small molecules. Both have similar size, but CO2's lack of polarity gives it an advantage in crossing the lipid core. A tempting distractor is urea because it is small (choice A), but this ignores the misconception that size alone determines speed, whereas polarity significantly reduces permeability in hydrophobic environments. To analyze similar problems, evaluate both size and polarity, prioritizing nonpolar molecules for faster diffusion across bilayers without proteins.

Question 20

1. Ca$^{2+}$, because it is slightly smaller and therefore crosses faster
2. CO$_2$, because nonpolar molecules partition into the bilayer interior (correct answer)
3. Ca$^{2+}$, because its charge allows it to pass between phospholipids
4. Both equally, because their molecular masses are similar
5. Neither, because membranes are impermeable to molecules near 40 Da

Explanation: This question tests the skill of analyzing membrane permeability based on solute properties in a phospholipid bilayer. CO₂ shows greater permeability because it is nonpolar, partitioning into the bilayer interior easily compared to the charged Ca²⁺. Ca²⁺ is repelled by the hydrophobic core, resulting in very low diffusion rates. The stimulus notes similar masses and no proteins, emphasizing polarity over size. A tempting distractor is choice A, favoring slight size difference, but this ignores the misconception that charge allows passage like nonpolarity. For transferable strategy, always rank nonpolar molecules above charged ones, regardless of minor mass similarities in bilayers.