Membrane Permeability — AP Biology Study Guide

For: AP Biology candidates sitting AP Biology.

Covers: Selective permeability of the phospholipid bilayer, factors altering membrane permeability, water potential relationships for permeable/impermeable solutes, and experimental analysis of permeability from lab data. Prepares for both MCQ and FRQ question types.

You should already know: Basic structure of the phospholipid bilayer, core principles of diffusion and osmosis, how to calculate water potential for solutions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Biology style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Membrane Permeability?

Membrane permeability is defined as the ability of a given solute or solvent to cross a biological phospholipid bilayer membrane. It is a continuous, quantitative property, not a binary "can/cannot cross" classification, though AP Biology typically frames it in terms of relative permeability for different molecule types. The selective permeability of cell membranes is a foundational concept for Unit 2: Cell Structure and Function, which accounts for 10-13% of the total AP Biology exam score, and membrane permeability questions appear regularly in both multiple-choice (MCQ) and free-response (FRQ) sections. Synonyms used in exam questions include "selective permeability" (the specific term for the cell membrane's property of allowing only certain substances to cross) and "diffusion permeability". The standard notation for quantitative permeability is the permeability coefficient $P$, with units of length per time (e.g., cm/s), though most AP problems test conceptual understanding and application to experimental scenarios rather than calculation of $P$ itself. This property underpins all cellular exchange with the environment, from nutrient uptake to waste removal to osmoregulation.

2. Selective Permeability by Molecule Class

The most fundamental rule of membrane permeability is that the interior of the phospholipid bilayer is hydrophobic (nonpolar), so permeability depends primarily on two properties of the crossing molecule: (1) polarity and charge, (2) molecular size. Nonpolar, small, uncharged molecules have the highest permeability, because they can readily dissolve in the hydrophobic core of the bilayer. Polar molecules are repelled by the nonpolar core, so they have low permeability unless they are very small. Charged molecules (ions) have extremely low permeability regardless of size, because their charge creates a thick hydration shell of water that makes them too hydrophilic and large to cross the hydrophobic core.

The standard ranking of permeability in a pure protein-free phospholipid bilayer (from highest to lowest) is: small nonpolar molecules (O₂, CO₂, steroid hormones) > small uncharged polar molecules (water, glycerol, urea) > large uncharged polar molecules (glucose, sucrose) > charged ions (Na⁺, K⁺, Cl⁻). Whole cell membranes contain embedded channel and transport proteins that dramatically increase permeability for low-permeability molecules, allowing cells to take up nutrients like glucose and regulate ion concentrations.

Worked Example

Rank the following molecules from highest membrane permeability to lowest in a pure phospholipid bilayer (no protein channels): glucose, O₂, Na⁺ ion, glycerol. Justify your ranking.

1. First, categorize each molecule by polarity, charge, and size: O₂ is small and nonpolar, glycerol is small and uncharged polar, glucose is large and uncharged polar, Na⁺ is a charged ion.
2. Highest permeability belongs to small nonpolar O₂: it dissolves easily in the hydrophobic bilayer interior, so it crosses rapidly.
3. Next is small uncharged polar glycerol: it is polar, so it crosses more slowly than O₂, but its small size allows it to pass through gaps between phospholipids at measurable rates.
4. Next is large uncharged polar glucose: its large size and polar nature make it very slow to cross the pure bilayer, so permeability is much lower than glycerol.
5. Lowest permeability is charged Na⁺: the charge creates a strong hydration shell that prevents it from dissolving in the hydrophobic core, so permeability is nearly zero.
6. Final ranking: ${\text{O}}_2>\text{glycerol}>\text{glucose}>{\text{Na}}^{+}$

Exam tip: Always check if the question specifies "pure phospholipid bilayer" (no proteins) or "cell membrane" (includes proteins). If proteins are present, permeability for ions and glucose will be much higher due to channels and transporters.

3. Factors Altering Membrane Permeability

The permeability of a biological membrane is not fixed; it changes based on the structural composition of the bilayer and environmental conditions. Three key factors tested regularly on the AP exam are temperature, fatty acid tail saturation, and cholesterol content.

• Temperature: Increasing temperature increases the kinetic energy of phospholipid molecules, increasing the average space between tails and raising permeability for all solutes. Decreasing temperature reduces kinetic energy, causes tighter packing, and lowers permeability. Only extreme high temperatures (far above physiological range) disrupt the bilayer structure entirely, causing massive leakage.
• Fatty acid saturation: Saturated fatty acids have straight tails that pack tightly together, reducing space between them and lowering permeability. Unsaturated fatty acids have kinked tails from double bonds that prevent tight packing, increasing space and raising permeability.
• Cholesterol: Cholesterol acts as a permeability buffer, with opposite effects depending on temperature. At high temperatures, cholesterol restricts phospholipid movement, reduces spacing between tails, and lowers permeability. At low temperatures, cholesterol prevents phospholipids from packing tightly, maintaining permeability and preventing the membrane from becoming too rigid.

Worked Example

Two pure phospholipid bilayers are prepared: Bilayer A is made of 100% saturated 16-carbon fatty acids, Bilayer B is made of 100% unsaturated 16-carbon fatty acids. Both are held at 25°C, and permeability to glycerol is measured. Which bilayer has higher permeability, and what is the effect of increasing temperature by 15°C on both bilayers?

1. Recall the effect of saturation on packing: saturated fatty acids have straight tails that pack tightly, while unsaturated fatty acids have kinked tails that cannot pack tightly.
2. Tighter packing reduces space between phospholipids, which reduces permeability. So Bilayer B (unsaturated) has looser packing and higher initial permeability to glycerol than Bilayer A.
3. Increasing temperature by 15°C (a moderate change that does not disrupt the bilayer) increases the kinetic energy of phospholipids in both bilayers, increasing the average distance between fatty acid tails.
4. Increased spacing reduces the barrier to glycerol crossing, so permeability to glycerol increases in both bilayers.
5. The relative difference in permeability remains: Bilayer B will still have higher permeability than Bilayer A at the higher temperature.

Exam tip: Remember that cholesterol has a dual effect depending on temperature. If the question does not specify temperature, cholesterol acts to maintain stable membrane permeability (buffer it against temperature changes), which is the core function you will be expected to state on the exam.

4. Experimental Analysis of Membrane Permeability

A common AP Biology lab experiment tests membrane permeability by using dialysis tubing (a synthetic membrane with similar permeability properties to cell membranes) or plant tissue to measure solute and water movement. The core principle is that net water movement follows water potential, which depends on the permeability of the solutes:

• Permeable solutes can cross the membrane, so they move down their concentration gradient until their concentration is equal on both sides at equilibrium.
• Impermeable solutes cannot cross, so they remain on their original side and create a permanent solute potential difference that drives net water movement.

Water potential is calculated as: $$\Psi ={\Psi }_s+{\Psi }_p$$ where ${\Psi }_s$ (solute potential) = $-iCRT$ and ${\Psi }_p$ (pressure potential) is 0 for an open beaker. Net water movement always occurs from an area of higher (less negative) water potential to lower (more negative) water potential.

Worked Example

A dialysis bag is permeable to water and monosaccharides, but impermeable to disaccharides. The bag is filled with 0.2 M sucrose (disaccharide) and 0.1 M glucose (monosaccharide), then placed in a beaker containing 0.1 M sucrose and 0.2 M glucose. What is the expected change in mass of the bag before equilibrium, and what are the solute concentrations at equilibrium (assume equal volumes inside and outside the bag)?

1. First categorize solutes by permeability: sucrose (disaccharide) is impermeable, glucose (monosaccharide) is permeable.
2. At equilibrium, permeable glucose will equalize its concentration across the membrane. Total glucose in the system is 0.1 M (inside) + 0.2 M (outside) split equally across two equal volumes, so glucose concentration becomes 0.15 M on both sides.
3. Impermeable sucrose remains on its original side: 0.2 M inside the bag, 0.1 M outside the bag.
4. Calculate total solute concentration at equilibrium: inside = 0.2 M sucrose + 0.15 M glucose = 0.35 M; outside = 0.1 M sucrose + 0.15 M glucose = 0.25 M.
5. Solute potential inside is ${\Psi }_{s(in)}=-0.35RT$, which is more negative (lower water potential) than outside ${\Psi }_{s(out)}=-0.25RT$. Net water moves into the bag, so the mass of the bag increases.
6. Final concentrations: 0.15 M glucose (both sides), 0.2 M sucrose (inside), 0.1 M sucrose (outside).

Exam tip: Always categorize solutes by permeability before calculating water potential. Do not add up initial total concentrations and assume they will equalize—impermeable solutes stay on their original side, which changes the final water potential difference.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Claiming cholesterol always decreases membrane permeability. Why: Students memorize that cholesterol restricts phospholipid movement, but forget the effect depends on environmental temperature. Correct move: Always specify the temperature context: cholesterol decreases permeability at high temperatures and increases permeability at low temperatures, acting as a permeability buffer.
• Wrong move: Ranking glucose as lower permeability than Na+ in a whole cell membrane. Why: Students confuse pure protein-free bilayers with whole cell membranes that contain transporters. Correct move: Always adjust rankings based on the presence of proteins specified in the question; in most cell membranes, glucose permeability is much higher than resting Na+ permeability due to glucose transporters.
• Wrong move: Calculating total initial solute concentration to predict mass change without separating permeable and impermeable solutes. Why: Students assume all solutes will equalize, missing that impermeable solutes stay on their original side. Correct move: First categorize each solute as permeable or impermeable, calculate equilibrium concentrations for permeable solutes, then sum total solute to find the water potential difference.
• Wrong move: Claiming large nonpolar molecules have higher permeability than small polar molecules. Why: Students only remember polarity and forget size as a determining factor. Correct move: Always evaluate both polarity/charge and size: a large nonpolar molecule has lower permeability than small nonpolar O₂, but higher permeability than small charged ions.
• Wrong move: Assuming any increase in temperature denatures membrane proteins and decreases permeability. Why: Students confuse moderate temperature increases with extreme temperature increases. Correct move: Only extreme high temperature (far above physiological range) denatures proteins and disrupts the bilayer; moderate temperature increases increase permeability by increasing phospholipid movement.

6. Practice Questions (AP Biology Style)

Question 1 (Multiple Choice)

Aquaporins are channel proteins that increase the permeability of cell membranes specifically to water. Which of the following describes the effect of adding aquaporins to a pure phospholipid bilayer placed in a hypertonic solution? A) The rate of net water movement out of the bilayer increases, and the equilibrium mass of the bilayer does not change B) The rate of net water movement out of the bilayer increases, and the equilibrium mass of the bilayer is lower C) The rate of net water movement out of the bilayer does not change, and the equilibrium mass of the bilayer is lower D) The rate of net water movement out of the bilayer decreases, and the equilibrium mass of the bilayer does not change

Worked Solution: Aquaporins increase membrane permeability to water, so they increase the rate of water movement across the membrane, which eliminates options C and D. Equilibrium is reached when water potential is equal on both sides of the membrane. Aquaporins only change the rate of water movement, not the final equilibrium position, because they do not alter the solute concentrations or the final water potential difference. The equilibrium mass depends only on the solute concentrations, which do not change when adding aquaporins. The correct answer is A.

Question 2 (Free Response)

A researcher investigates the effect of fatty acid saturation on the permeability of beetroot cell membranes to betacyanin, a red pigment found in beetroot vacuoles. The researcher cuts equal-sized beetroot disks, places them in 45°C water baths for 10 minutes, then measures the absorbance of red light in the surrounding water (higher absorbance = more betacyanin leaked out = higher membrane permeability). The results are below:

(a) Describe the relationship between percent unsaturated fatty acids and permeability, and justify your conclusion using the data. (b) A fourth variety has 65% unsaturated fatty acids. Predict the absorbance expected for this variety, and show your reasoning. (c) Researchers add additional cholesterol to the membranes of variety 3 held at 45°C. Predict the change in absorbance, and connect this to the role of cholesterol.

Worked Solution: (a) There is a positive correlation between percent unsaturated fatty acids and membrane permeability: as unsaturated fatty acid content increases, absorbance increases, meaning more betacyanin leaks out, so permeability increases. This matches the expected effect of unsaturated fatty acids: kinked tails from double bonds prevent tight packing of phospholipids, increasing spacing between tails and making it easier for betacyanin to leak out of the vacuole. (b) The trend shows that a 30% increase in unsaturated fatty acids (from 20% to 50%) leads to a 0.22 increase in absorbance, which is ~0.0073 absorbance units per 1% unsaturated fatty acid. A 15% increase from 50% to 65% gives an expected absorbance increase of ~0.11, so total absorbance ≈ 0.34 + 0.11 = 0.45. Any value between 0.4 and 0.5 that follows the positive trend is acceptable. (c) 45°C is a high temperature relative to the normal physiological temperature of beetroot. At high temperatures, cholesterol restricts phospholipid movement, reduces spacing between tails, and decreases membrane permeability. Lower permeability means less betacyanin leaks out, so absorbance will decrease compared to the original 0.34.

Question 3 (Application / Real-World Style)

Antarctic ice fish live in seawater that is constantly at -1.8°C, well below the physiological temperature of most tropical fish. To maintain normal membrane permeability at this low temperature, ice fish have evolved modified membrane fatty acid and cholesterol composition compared to tropical fish. Predict the difference in (1) fatty acid saturation and (2) cholesterol content between ice fish and tropical fish membranes, and explain why this adaptation is necessary for survival.

Worked Solution:

1. Low temperature reduces the kinetic energy of phospholipids, causing them to pack tightly together, which decreases membrane permeability below the level required for normal cellular nutrient and waste exchange.
2. (1) Ice fish will have a higher proportion of unsaturated fatty acids in their membranes than tropical fish. The kinked tails of unsaturated fatty acids prevent tight packing at low temperatures, maintaining sufficient spacing between phospholipids to keep permeability at functional levels.
3. (2) Ice fish will also have higher cholesterol content than tropical fish. At low temperatures, cholesterol prevents phospholipids from packing tightly together, preserving membrane permeability and preventing the membrane from becoming too rigid.
4. In context: This adaptation allows ice fish to maintain normal cellular exchange across cell membranes at the near-freezing temperatures of their environment, which would otherwise cause membranes to become too impermeable to support life.

7. Quick Reference Cheatsheet

8. What's Next

Membrane permeability is the foundational prerequisite for understanding transport across cell membranes, the next core topic in Unit 2: Cell Structure and Function. Without understanding which molecules can cross the membrane and what factors alter permeability, you cannot correctly predict the movement of solutes and water via passive and active transport, a major focus of AP Biology FRQs. Beyond Unit 2, membrane permeability underpins concepts like cell signaling (ligands must cross the membrane or bind to surface receptors), osmoregulation in whole organisms, and cell compartmentalization, all of which rely on the selective permeability of internal and external cell membranes. Mastery of this topic also makes it easier to analyze experimental data on membrane transport, a common skill tested on the exam.

• Passive and Active Transport
• Water Potential and Osmosis
• Membrane Structure
• Cell Compartmentalization