Membrane Permeability
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AP Biology › Membrane Permeability
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?
Both, because neither molecule is charged
Isopropanol, because it is less polar and smaller than ribose
Ribose, because it has more oxygen atoms to interact with the membrane
Isopropanol, because polar molecules cross faster than nonpolar molecules
Ribose, because sugars are used by cells and therefore diffuse easily
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.
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?
Ethanol is uncharged, so it is excluded by the membrane surface
Propane is larger, so it is pushed through by collisions more often
Propane is nonpolar, so it dissolves in the hydrophobic core more easily
Propane crosses faster because water repels it into the membrane
Ethanol is polar, so it must use ATP to cross any 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.
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.
Both cross at similar rates, because equal external concentration eliminates permeability differences.
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.
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.
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
Glyceraldehyde arrives first because it is larger and moves down gradients faster
Neither arrives because uncharged molecules cannot cross a bilayer
Both arrive equally because diffusion depends only on concentration difference
O$_2$ arrives first because nonpolar molecules cross the hydrophobic core readily
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.
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?
Argon, because nonpolar molecules pass readily through the hydrophobic core
Both equally, because they have nearly the same molecular mass
K$^+$, because it is slightly smaller and therefore diffuses more rapidly
Neither, because diffusion across membranes requires ATP hydrolysis
K$^+$, because charged particles are attracted to phospholipid tails
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.
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?
Methane, because nonpolar molecules cannot dissolve in water
Both, because neither is charged
Methane, because it is small and nonpolar
Glycerol, because it is smaller than most sugars
Glycerol, because its hydroxyl groups interact with the membrane surface
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.
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$_3$ (17 Da, uncharged, polar)
N$_2$ (28 Da, nonpolar)
NH$_4^+$ (18 Da, charged)
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.
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 positive charge attracts it to the membrane
Ammonium (NH4+), because it is small and contains hydrogen
Both, because nitrogen-containing molecules cross membranes readily
Nitrous oxide (N2O), because it is uncharged and relatively nonpolar
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.
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 charges are attracted to lipid tails
Both, because they are derived from the same amino acid
Alanine methyl ester, because esters are always charged in water
Alanine methyl ester, because it is neutral and less polar
Alanine (zwitterion), because it is small and can hydrogen-bond
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.
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?
Neither, because gases cannot cross a liquid membrane
Both, because they are made of the same element
Chlorine gas (Cl2), because it is nonpolar and uncharged
Chloride ion (Cl−), because it is smaller than Cl2
Chloride ion (Cl−), because ions dissolve in water and thus enter membranes
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.