- A membrane consists of a phospholipid bilayer with no proteins. The hydrophobic interior strongly limits passage of polar and charged molecules. Consider two solutes: formamide (small, strongly polar due to a carbonyl and amine) and carbon monoxide (CO), which is small and relatively nonpolar. Both are uncharged. Temperature and concentration gradients are held constant.
Which solute would most likely have greater permeability across the bilayer?
Answer: Carbon monoxide (CO), because it is small and relatively nonpolar
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.
- A sealed artificial phospholipid bilayer separates two solutions. Five solutes are added separately at equal concentration: O2 (32 Da, nonpolar), ethanol (46 Da, polar uncharged), glucose (180 Da, polar uncharged), Na+ (23 Da, charged), and a 12–amino acid peptide (~1300 Da, polar). No transport proteins are present. Which solute would most likely cross the bilayer fastest by simple diffusion?
Answer: O2 (32 Da, nonpolar)
This question tests the skill of analyzing membrane permeability based on solute properties in a phospholipid bilayer. O₂ crosses the bilayer fastest because it is small and nonpolar, allowing it to easily dissolve in and diffuse through the hydrophobic core of the membrane without needing transport proteins. In contrast, ethanol and glucose are polar uncharged, which hinders their entry into the nonpolar interior, while Na⁺ is charged and strongly repelled, and the large peptide faces both size and polarity barriers. The stimulus details emphasize equal concentrations and no proteins, highlighting that simple diffusion favors nonpolar solutes like O₂ over others. A tempting distractor is Na⁺ (choice D), as its small size might suggest fast diffusion, but this ignores the misconception that charge prevents ions from crossing the hydrophobic barrier effectively. For transferable strategy, always prioritize nonpolarity over size for diffusion through pure lipid bilayers, then consider molecular weight for polar uncharged solutes. - A pure phospholipid bilayer is exposed to equal concentrations of two 100 Da solutes: solute R is nonpolar; solute S is polar uncharged. No proteins are present. Which explanation best accounts for the expected difference in their permeability?
Answer: Solute R crosses faster because nonpolarity increases solubility in lipid tails
This question tests the skill of analyzing membrane permeability based on solute properties in a phospholipid bilayer. Solute R crosses faster because its nonpolarity increases solubility in lipid tails, unlike the polar uncharged solute S. Equal masses isolate polarity's effect on permeability. The stimulus specifies 100 Da and no proteins, emphasizing nonpolar advantage. A tempting distractor is choice A, claiming polar interacts with core, but this ignores the misconception that polarity aids nonpolar dissolution. For transferable strategy, always favor nonpolar over polar when sizes match, as lipid solubility drives diffusion rates.
- 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?
Answer: Oxygen, because nonpolar molecules enter the hydrophobic interior more readily
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.
- 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?
Answer: Propane is nonpolar, so it dissolves in the hydrophobic core more easily
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 phospholipid bilayer lacking proteins is exposed to equal concentrations of iodine (I2, 254 Da, nonpolar) and lysine (146 Da, charged at neutral pH). Which molecule would most likely cross the membrane more readily by simple diffusion?
Answer: I2, because nonpolar molecules can dissolve in the hydrophobic bilayer core
This question tests the skill of analyzing membrane permeability based on solute properties in a phospholipid bilayer. I₂ crosses more readily because it is nonpolar, dissolving in the hydrophobic core unlike the charged lysine. Despite I₂'s larger mass, nonpolarity overcomes size barriers. The stimulus provides masses and properties with no proteins, underscoring polarity's dominance. A tempting distractor is choice A, favoring lysine's smaller size, but this ignores the misconception that charge allows stabilization in bilayers. For transferable strategy, always rank nonpolar over charged solutes, even if larger, for bilayer permeability. - A phospholipid bilayer separates two solutions. The bilayer interior is nonpolar and resists entry of molecules that are polar or charged. Compare two molecules of similar size: acetone (polar due to a carbonyl group) and hexane (nonpolar hydrocarbon). Neither molecule is ionized in this context. The membrane contains no proteins.
Which molecule would most likely have greater permeability through the bilayer?
Answer: Hexane, because it is nonpolar and partitions into the membrane
This question assesses the skill of analyzing membrane permeability based on molecular properties in a phospholipid bilayer. Hexane has greater permeability than acetone because it is a nonpolar hydrocarbon, partitioning easily into the nonpolar interior, while acetone's carbonyl group makes it polar and less soluble. The similar size emphasizes that nonpolarity is key for dissolving in the hydrophobic region. Without proteins, simple diffusion relies on lipid solubility, favoring hexane. A tempting distractor is acetone because it has fewer atoms (choice C), but this reflects the misconception that smaller atom count always means higher permeability, ignoring polarity's role in reducing it. To analyze similar problems, prioritize nonpolar molecules over polar ones of similar size when evaluating bilayer permeability without transporters.
- A phospholipid bilayer has a hydrophobic interior and hydrophilic surfaces. Consider five solutes at equal concentration: O2 (nonpolar), CO2 (nonpolar), H2O (small polar), glucose (large polar), and Na+ (small ion). None has a transport protein available. Molecules cross only by simple diffusion through the lipid portion of the membrane. Permeability depends on how well a solute dissolves in the hydrophobic core and how strongly it interacts with water. Charged solutes remain strongly hydrated and interact poorly with the nonpolar interior. Larger polar solutes form many hydrogen bonds with water, reducing partitioning into the membrane.
Answer: O2, because nonpolar molecules partition into the hydrophobic core most readily
This question tests your ability to analyze membrane permeability based on molecular properties. The phospholipid bilayer has a hydrophobic interior that favors nonpolar molecules, while polar and charged molecules face energetic barriers to entry. O2 is nonpolar and can readily dissolve in the hydrophobic core, making it the most permeable molecule among the choices. In contrast, Na+ (choice A) is incorrect because ions remain strongly hydrated and cannot shed their water shell to enter the nonpolar interior—small size alone doesn't overcome the charge barrier. When comparing membrane permeability, evaluate both polarity and charge: nonpolar > polar > charged molecules.
- Two solutes are compared for diffusion through a protein-free phospholipid bilayer: solute P (propane, 44 Da, nonpolar) and solute Q (serine, 105 Da, polar with a zwitterionic charge). Which solute would most likely cross the membrane faster?
Answer: Propane, because small nonpolar molecules pass readily through lipid tails
This question tests the skill of analyzing membrane permeability based on solute properties in a phospholipid bilayer. Propane crosses faster because it is small and nonpolar, passing readily through the lipid tails unlike the polar, zwitterionic serine. Serine's charge and polarity create strong barriers to entering the hydrophobic core. The stimulus compares sizes and properties with no proteins, underscoring nonpolarity's advantage. A tempting distractor is choice B, claiming amino acids enter easily, but this reflects the misconception that biological relevance overrides physical properties. For transferable strategy, always favor small nonpolar over polar charged solutes, ignoring functional categories in diffusion analysis.
- A pure phospholipid bilayer is tested with two uncharged solutes: solute M (40 Da, polar) and solute N (400 Da, nonpolar). No proteins are present. Which prediction is most consistent with membrane permeability rules?
Answer: Solute N crosses faster because nonpolarity strongly favors passage despite larger size
This question tests the skill of analyzing membrane permeability based on solute properties in a phospholipid bilayer. Solute N crosses faster because its nonpolarity strongly favors passage through the hydrophobic core, despite being much larger than the polar solute M. Solute M's polarity reduces its lipid solubility, limiting diffusion even with small size. The stimulus notes uncharged nature and no proteins, highlighting nonpolarity's dominance over size differences. A tempting distractor is choice A, claiming small size always wins, but this ignores the misconception that polarity barriers are secondary to mass. For transferable strategy, always evaluate nonpolarity as outweighing moderate size increases for bilayer permeability.
- A phospholipid bilayer separates two solutions. Two solutes are tested for simple diffusion through the lipid portion only: NH3 (ammonia; small, uncharged) and NH4+ (ammonium; small, positively charged). The solutions have the same total nitrogen concentration, but the membrane is impermeable to charged species unless they can enter the hydrophobic core. Charged solutes remain strongly associated with water molecules, making entry into the nonpolar interior unfavorable. The uncharged form has fewer electrostatic interactions with water and can partition into the bilayer more readily.
Answer: NH3, because the uncharged form more readily enters the hydrophobic interior
This question examines membrane permeability differences between charged and uncharged forms of the same molecule. The phospholipid bilayer's hydrophobic core strongly excludes charged species because they remain tightly associated with water molecules through electrostatic interactions. NH3 is uncharged and can partition into the nonpolar interior, while NH4+ carries a positive charge that keeps it strongly hydrated and prevents membrane entry. Choice A incorrectly assumes that positive charges are attracted to negative phosphate groups, but phosphate groups face outward toward water, not into the hydrophobic core where diffusion occurs. For weak acids and bases, remember that only the uncharged form can cross membranes by simple diffusion—charge creates an insurmountable barrier.
- A pure phospholipid bilayer is exposed to equal concentrations of Ca2+ (40 Da, charged) and CO2 (44 Da, nonpolar). No proteins are present. Which molecule would most likely show greater permeability across the membrane?
Answer: CO2, because nonpolar molecules partition into the bilayer interior
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. - A pure phospholipid bilayer separates two aqueous solutions. Molecules can cross only by simple diffusion through the hydrophobic core (no proteins present). Consider these molecules at similar concentrations: O2 (nonpolar, small), glucose (large, polar), Na+ (small, charged), and ethanol (small, polar uncharged). Because the bilayer interior is nonpolar, molecules that are nonpolar and small dissolve into the core more readily than polar or charged molecules. Charged ions remain strongly associated with water and are energetically unfavorable within the hydrophobic region. Larger molecules, even if uncharged, have more difficulty fitting between phospholipids than smaller molecules.
Answer: O2, because small nonpolar molecules partition into the hydrophobic core most readily.
This question assesses the skill of analyzing membrane permeability based on molecular properties in a pure phospholipid bilayer. The correct answer is O₂ because it is small and nonpolar, allowing it to readily partition into and diffuse through the hydrophobic core of the bilayer, as described in the stimulus where nonpolar molecules dissolve more easily than polar or charged ones. In contrast, glucose is large and polar, making it less able to fit between phospholipids and less soluble in the nonpolar interior, while Na⁺ is charged and remains associated with water, rendering its passage energetically unfavorable. Ethanol, though small and uncharged, is polar, so it diffuses slower than nonpolar O₂ despite some ability to cross. A tempting distractor is choice A, which incorrectly suggests that Na⁺'s small size overcomes its charge, reflecting the misconception that size alone determines permeability without considering charge interactions with the hydrophobic region. A transferable strategy is to always evaluate both polarity and size when predicting simple diffusion rates across lipid bilayers. - A synthetic membrane is made of phospholipids and has no embedded proteins. Two solutes are placed on one side at equal concentrations: ethanol (2 carbons, one hydroxyl; uncharged) and ethylene glycol (2 carbons, two hydroxyls; uncharged). The bilayer interior is nonpolar, so solutes must leave their hydration shell to enter it. Molecules with more hydroxyl groups form more hydrogen bonds with water and are more strongly hydrated. Because ethanol and ethylene glycol are similar in size, the primary determinant of permeability is polarity rather than molecular mass.
Answer: Ethanol, because fewer hydrogen bonds with water increase partitioning into the bilayer
This problem tests membrane permeability analysis based on the number of polar groups. The synthetic membrane's nonpolar interior requires solutes to leave their hydration shell to enter, which is energetically costly for polar molecules. Ethanol has one hydroxyl group while ethylene glycol has two, making ethylene glycol more polar and more strongly hydrogen-bonded to water molecules. Therefore, ethanol can more readily partition into the bilayer and cross faster. Choice A incorrectly suggests that additional hydroxyl groups increase attraction to membrane lipids, when they actually increase attraction to water and decrease membrane permeability. When comparing molecules of similar size, count hydroxyl groups—each additional OH group decreases membrane permeability by increasing water interactions.
- A pure phospholipid bilayer is used to compare diffusion of two steroids: cortisol (has several polar hydroxyl groups) and cholesterol (fewer polar groups, largely nonpolar). Both are large and uncharged, but cortisol is more polar. The bilayer interior is hydrophobic, so increased polarity reduces solubility in the membrane core. No transport proteins are present.
Which molecule would most likely have higher permeability through the bilayer?
Answer: Cholesterol, because it is less polar and more lipid-soluble
This question assesses the skill of analyzing membrane permeability based on molecular properties in a phospholipid bilayer. Cholesterol has higher permeability than cortisol because it is less polar and more lipid-soluble with fewer polar groups, allowing better partitioning into the hydrophobic core, while cortisol's hydroxyl groups increase polarity. Both are large uncharged steroids, but cholesterol's greater nonpolarity enhances diffusion. The pure bilayer without proteins favors lipid-like molecules. A tempting distractor is cortisol because it is a hormone (choice C), but this is due to the misconception that function affects physical permeability, whereas molecular properties determine it. To analyze similar problems, compare polarity in related compounds, choosing the least polar for higher bilayer permeability.
- 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?
Answer: Nitrous oxide (N2O), because it is uncharged and relatively nonpolar
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 membrane is composed of a phospholipid bilayer with a hydrophobic interior. Two solutes are placed on one side: benzoic acid in its uncharged form (C6H5COOH; weak acid, uncharged) and benzoate (C6H5COO−; negatively charged). No transport proteins are present, so crossing requires partitioning into the lipid core. Although both forms have similar size, the charged form remains strongly hydrated and interacts poorly with the nonpolar interior. The uncharged form has reduced electrostatic interactions with water and can enter the bilayer more readily by diffusion.
Answer: Benzoic acid (uncharged), because lack of charge increases lipid solubility and diffusion
This question examines membrane permeability for weak acids in charged versus uncharged forms. The phospholipid bilayer's hydrophobic core strongly excludes charged molecules because they remain hydrated through electrostatic interactions with water. Benzoic acid in its uncharged form can partition into the nonpolar interior, while the negatively charged benzoate ion cannot shed its hydration shell to enter the membrane. This principle explains why weak acids can cross membranes more readily at low pH where they are protonated and uncharged. Choice A incorrectly claims that negative charge increases attraction to the hydrophobic core, when charge actually prevents entry into this nonpolar region. Remember that for ionizable molecules, only the uncharged form crosses membranes by simple diffusion.
- D-glucose uptake saturates without ATP; L-glucose uptake is linear with concentration. This shows that
Answer: D-glucose uses a carrier
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.
- A protein-free phospholipid bilayer separates two solutions. Equal concentrations of H+ (1 Da, charged), glucose (180 Da, polar uncharged), and CO2 (44 Da, nonpolar) are added to one side. Which solute would most likely show the greatest net movement across the bilayer over a short time?
Answer: CO2, because nonpolar molecules cross the hydrophobic interior readily
This question tests the skill of analyzing membrane permeability based on solute properties in a phospholipid bilayer. CO₂ shows the greatest net movement because it is nonpolar, crossing the hydrophobic interior readily unlike charged H⁺ or polar glucose. H⁺ is impermeable due to charge, and glucose is slow due to size and polarity. The stimulus adds equal concentrations to one side with no proteins, focusing on short-term diffusion. A tempting distractor is choice A (H⁺), due to tiniest size, but this reflects the misconception that charge doesn't block tiny ions. For transferable strategy, always identify nonpolar solutes for fastest net flux, considering barriers for charged and polar ones. - Red cells lyse in 0.3 M urea but not in 0.15 M NaCl. Why?
Answer: Urea crosses; water follows
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.
