Membrane Transport — AP Biology Study Guide

For: AP Biology candidates sitting AP Biology.

Covers: Selective permeability, passive transport, osmosis, water potential calculations, tonicity, active transport, electrochemical gradients, cotransport, and bulk transport (endocytosis/exocytosis). This chapter aligns with the 2024 AP Biology CED learning objectives.

You should already know: Fluid mosaic model structure of the phospholipid bilayer. Concentration gradients and basic chemical properties of polar/nonpolar molecules. Homeostasis as a core requirement for cell function.

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 Transport?

Membrane transport describes the movement of ions, molecules, and other substances across biological cell membranes, a process required to maintain cellular homeostasis by controlling the internal chemical environment of the cell. This topic is a core component of Unit 2: Cell Structure and Function, which accounts for 10–13% of the total AP Biology exam score, with membrane transport making up roughly one-third of that unit’s content (3–4% of the total exam). It appears in both multiple-choice questions (MCQ) and free-response questions (FRQ), often combined with other topics like cell size, osmoregulation, cell signaling, and energy metabolism. All membrane transport processes are governed by the selective permeability of the phospholipid bilayer: only small, nonpolar molecules can cross freely, while charged, polar, or large molecules require specialized proteins or energy input to move across the membrane.

2. Selective Permeability and Passive Transport

Selective permeability is the membrane’s ability to restrict movement of some molecules while allowing others to cross, a property arising directly from the structure of the phospholipid bilayer. The hydrophobic core of the bilayer, formed by the fatty acid tails of phospholipids, repels charged and polar molecules, while allowing nonpolar molecules to dissolve and diffuse freely. Passive transport is any movement of molecules across the membrane that occurs down a concentration gradient (from higher to lower concentration) and does not require input of cellular energy (ATP). There are two types of passive transport: (1) simple diffusion, where molecules move directly across the bilayer without assistance; and (2) facilitated diffusion, where molecules move through integral membrane proteins (channels or carriers) because they cannot cross the hydrophobic core on their own. Channel proteins form open pores for specific molecules (e.g., aquaporins for water), while carrier proteins change shape to shuttle larger molecules like glucose across the membrane. Both types of passive transport are driven by the kinetic energy of the molecules themselves, not ATP.

Worked Example

A researcher reconstitutes an artificial phospholipid bilayer (no integral proteins) and measures the relative rate of movement of three molecules: oxygen (O₂), glucose, and sodium ion (Na⁺). Predict the order of movement from fastest to slowest and justify your prediction.

1. First, categorize each molecule by size and chemical properties: O₂ is a small nonpolar molecule, glucose is a large polar molecule, and Na⁺ is a charged ion.
2. The hydrophobic core of the bilayer only allows nonpolar molecules to dissolve and diffuse freely; it strongly repels charged molecules and blocks large polar molecules from crossing at appreciable rates.
3. O₂ can diffuse directly through the bilayer, so it has the fastest rate. Glucose is too large and polar to cross quickly, so it has an intermediate rate. Na⁺ charge makes it unable to cross the hydrophobic core at all without a channel, so it has the slowest rate.
4. Final order (fastest to slowest): O₂ > glucose > Na⁺.

Exam tip: On FRQ justifications for permeability, always link the molecule’s properties to the structure of the bilayer explicitly—AP readers require this connection to award full points, not just a ranking of rates.

3. Water Potential and Osmosis

Osmosis is the net diffusion of free water across a selectively permeable membrane, and its direction is predicted using water potential ($\Psi$, psi), a measure of the potential energy of water to move. Water always moves from a region of higher water potential to a region of lower water potential. The formula for total water potential is: $$\Psi ={\Psi }_s+{\Psi }_p$$ where ${\Psi }_s$ = solute potential (also called osmotic potential) and ${\Psi }_p$ = pressure potential. Solute potential is always negative because adding solute reduces the number of free water molecules, lowering water potential; pure water has a solute potential of 0. The formula for solute potential is: $${\Psi }_s=-iCRT$$ where $i$ = ionization constant (number of particles a solute splits into when dissolved), $C$ = molar concentration of solute, $R$ = pressure constant ($0.0831L\cdot bar/mol\cdot K$), and $T$ = temperature in Kelvin (${}^{\circ }C+273$). Tonicity describes the effect of a solution on cell volume: hypotonic solutions have lower solute concentration than the cell, hypertonic have higher, and isotonic have equal solute concentration. Plant cells have a rigid cell wall that generates positive turgor pressure when water enters, so they are healthiest in hypotonic environments, while animal cells lack a cell wall and burst in hypotonic conditions.

Worked Example

Calculate the total water potential of a 0.20 M sucrose solution in an open beaker at 27°C. If a plant cell with $\Psi =-0.8bars$ is placed in this solution, what direction will water move?

1. Convert temperature to Kelvin: $27^{\circ }C+273=300K$.
2. Sucrose does not ionize in water, so $i=1$.
3. Calculate solute potential: ${\Psi }_s=-(1)(0.20mol/L)(0.0831L\cdot bar/mol\cdot K)(300K)=-4.99bars\approx -5.0bars$.
4. Open beakers have 0 pressure potential (atmospheric pressure), so total water potential of the solution is $\Psi =-5.0+0=-5.0bars$.
5. Compare water potentials: the plant cell has $\Psi =-0.8bars$ (higher than the solution’s $-5.0bars$). Water moves from higher to lower water potential, so net water movement is out of the plant cell into the solution.

Exam tip: Double-check the ionization constant $i$ before solving: use $i=1$ for non-electrolytes (sucrose, glucose) and $i=2$ for NaCl (and other 1:1 salts that dissociate into two ions). AP questions often test this common mistake.

4. Active Transport

Active transport is the movement of molecules across the membrane against their concentration gradient (from lower to higher concentration), which requires input of cellular energy (usually ATP) and is carried out by transmembrane protein pumps. The most well-studied example is the Na⁺/K⁺ ATPase pump, which pumps 3 Na⁺ ions out of the cell and 2 K⁺ ions into the cell, both against their gradients, using one ATP per cycle. This pump establishes and maintains an electrochemical gradient (a combined gradient of charge and concentration) across the membrane, which is critical for nerve function, nutrient uptake, and muscle contraction. Cotransport is a secondary active transport process: a proton pump first uses ATP to establish a H⁺ gradient, then the downhill diffusion of H⁺ provides the energy to move a second molecule (e.g., sucrose) uphill against its gradient. Unlike passive transport, active transport can create and maintain concentration differences that would never form via diffusion alone.

Worked Example

A researcher treats cultured mammalian cells with a toxin that inhibits ATP synthesis. After treatment, they observe that intracellular Na⁺ concentration increases and intracellular K⁺ concentration decreases, but facilitated diffusion of glucose across the membrane is unchanged. Explain this observation.

1. The Na⁺/K⁺ pump is an active transport protein that requires ATP to move Na⁺ out and K⁺ in against their concentration gradients.
2. Inhibiting ATP synthesis removes the energy source required for the pump to function, so it can no longer maintain the gradient. Na⁺ leaks down its gradient into the cell, and K⁺ leaks out, causing the observed concentration changes.
3. Facilitated diffusion of glucose is a passive process that moves glucose down its concentration gradient through a carrier protein, and does not require ATP.
4. Because no energy is needed for facilitated diffusion, it is unaffected by the loss of ATP, matching the researcher’s observation.

Exam tip: To earn full points when distinguishing active vs passive transport on FRQs, always explicitly state two key differences: (1) gradient direction (against vs down), and (2) energy requirement (ATP required vs no ATP required).

5. Bulk Transport

Bulk transport is the movement of large particles, macromolecules, or large volumes of extracellular fluid that cannot fit through transport proteins, and it requires energy for vesicle formation and movement, making it a form of active transport. It occurs via two main processes: endocytosis (movement into the cell) and exocytosis (movement out of the cell). Endocytosis has three subtypes: phagocytosis ("cell eating"), where the cell engulfs large particles like bacteria into a food vacuole; pinocytosis ("cell drinking"), where the cell takes up small droplets of extracellular fluid with dissolved solutes; and receptor-mediated endocytosis, a highly specific process where receptors on the cell surface bind target molecules (e.g., LDL cholesterol) before the membrane folds in to bring them into the cell. Exocytosis occurs when vesicles from the Golgi apparatus fuse with the cell membrane to release their contents outside the cell, for example, secretion of insulin from pancreatic cells or release of neurotransmitters from neurons.

Worked Example

The genetic condition familial hypercholesterolemia is caused by a mutation that eliminates functional LDL receptor proteins on the surface of human body cells. LDL carries cholesterol in the bloodstream, and LDL enters cells via receptor-mediated endocytosis. Predict the effect of this mutation on blood cholesterol levels and justify your prediction.

1. Receptor-mediated endocytosis requires specific receptor proteins to bind target molecules before endocytosis can occur.
2. Without functional LDL receptors, LDL cholesterol cannot bind to body cell surfaces and cannot be taken up into cells from the bloodstream.
3. Unabsorbed LDL remains in the bloodstream, leading to chronically elevated blood cholesterol levels.
4. High blood cholesterol leads to plaque buildup in arteries, which causes the cardiovascular symptoms of familial hypercholesterolemia.

Exam tip: Do not forget that bulk transport is a form of active transport—it requires energy, even though it does not use protein pumps. AP MCQs often test this common misconception.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Calculating solute potential for NaCl and using an ionization constant $i=1$. Why: Students generalize the $i=1$ rule used for sucrose to all solutes, forgetting that ionic salts dissociate into multiple particles in solution. Correct move: Always confirm the solute before calculating: use $i=1$ for non-ionizing solutes (sucrose, glucose) and $i=2$ for 1:1 salts like NaCl.
• Wrong move: Stating water moves from higher solute concentration to lower solute concentration during osmosis. Why: Students mix up solute concentration and water potential, reversing the direction of movement. Correct move: Always describe osmosis direction using water potential: water moves from higher $\Psi$ to lower $\Psi$, which is from lower solute concentration to higher solute concentration.
• Wrong move: Claiming facilitated diffusion is active transport because it uses proteins. Why: Students associate membrane proteins with active transport, forgetting that passive transport can also use proteins as long as movement is down the gradient. Correct move: Classify transport based on energy requirement and gradient direction, not whether a protein is involved.
• Wrong move: Stating plant cells burst in hypotonic solutions. Why: Students mix up the responses of animal and plant cells to tonicity, forgetting the rigid cell wall. Correct move: Remember animal cells burst in hypotonic solution (no cell wall to resist pressure); plant cells become turgid (healthy) in hypotonic solution, and plasmolyze in hypertonic solution.
• Wrong move: Assuming all polar molecules cannot cross the phospholipid bilayer. Why: Students overgeneralize the permeability rule to all polar molecules regardless of size. Correct move: Small polar molecules (water, urea) can cross the bilayer slowly, while large polar molecules (glucose, starch) require transport proteins to cross at biologically relevant rates.
• Wrong move: Stating the Na⁺/K⁺ pump moves equal numbers of Na⁺ and K⁺ across the membrane. Why: Students memorize the pump’s function but forget the unequal movement that creates the resting membrane potential. Correct move: Recall the 3 out, 2 in rule: 3 Na⁺ exit the cell, 2 K⁺ enter, for every 1 ATP hydrolyzed.

7. Practice Questions (AP Biology Style)

Question 1 (Multiple Choice)

Which of the following molecules will diffuse most quickly across an artificial phospholipid bilayer with no integral membrane proteins? A) H₂O B) O₂ C) Glucose D) Na⁺

Worked Solution: To answer this, we rank permeability based on the properties of the bilayer and each molecule. The hydrophobic core of the bilayer allows small nonpolar molecules to diffuse freely, blocks charged molecules completely, and slows small polar molecules. O₂ is small and nonpolar, so it diffuses fastest. H₂O is polar, so it diffuses much more slowly without aquaporins. Glucose is large and polar, and Na⁺ is charged, so both diffuse very slowly or not at all. The correct answer is B.

Question 2 (Free Response)

A biologist investigates osmosis in carrot cores. She cuts identical carrot cores, weighs them, and places each core in a different sucrose molarity at 25°C in open beakers. After 24 hours, she calculates percent change in mass, with results below:

(a) Estimate the molar concentration of solutes inside the carrot cells. Justify your estimate. (b) Calculate the total water potential of the carrot cells at equilibrium. Show all your work. (c) Predict what would happen to a human red blood cell placed in the 0.0 M sucrose solution. Explain your prediction.

Worked Solution: (a) No net water movement occurs when the water potential inside the cell equals the water potential outside, which corresponds to 0% change in mass. 0% change falls between 0.2 M (+1%) and 0.3 M (-3%), so the approximate solute concentration of carrot cells is 0.22 M. (b) Calculate solute potential: ${\Psi }_s=-iCRT=-(1)(0.22M)(0.0831L\cdot bar/mol\cdot K)(25+273)=-(0.22)(0.0831)(298)\approx -5.45bar$ At equilibrium, ${\Psi }_{carrot}={\Psi }_{solution}$. The solution is open, so ${\Psi }_p=0$, so total ${\Psi }_{carrot}=-5.45bar\approx -5.5bar$. (c) 0.0 M sucrose is pure water with $\Psi =0$, which is higher than the water potential inside a red blood cell ( -7 bar). Water moves into the red blood cell down the water potential gradient. Red blood cells have no cell wall, so they swell and eventually lyse (burst).

Question 3 (Application / Real-World Style)

A hiker becomes lost and runs out of clean drinking water, so they drink 1 L of seawater, which has a NaCl concentration of ~0.60 M. Calculate the total water potential of seawater at 20°C, and predict the effect of drinking seawater on the hiker’s red blood cells. Explain your reasoning.

Worked Solution: Convert temperature to Kelvin: $20+273=293K$. NaCl dissociates into two ions, so $i=2$. ${\Psi }_s=-(2)(0.60M)(0.0831)(293)=-(1.2)(24.35)\approx -29.2bar$. Seawater is open to the atmosphere, so ${\Psi }_p=0$, so total $\Psi =-29.2bar$. Human red blood cells have a total water potential of ~ -7.7 bar (isotonic to 0.15 M NaCl). Water moves from higher $\Psi$ inside red blood cells to lower $\Psi$ in the seawater outside (in the gut and bloodstream), so water leaves the red blood cells, causing them to shrink (crenate). In context, this crenation and increased blood osmolarity causes dehydration of cells and can lead to organ failure if not treated.

8. Quick Reference Cheatsheet

9. What's Next

Membrane transport is the foundational prerequisite for nearly all cellular processes covered later in the AP Biology course. Next, you will apply the concepts of gradients and selective permeability to chemiosmosis in cellular respiration and photosynthesis, where proton gradients across membranes drive ATP synthesis. Without mastering the rules of membrane transport and water potential, you will not be able to explain how energy is transferred in these core processes. Membrane transport also underlies all cell communication in Unit 4, where receptor localization and ligand movement depend on permeability properties. This topic also connects to osmoregulation in the physiology unit, where organisms maintain water balance across cell membranes.

• Cell Membrane Structure
• Cell Size and Surface Area
• Cellular Respiration
• Cell Communication