Lesson 3: Membranes and Cellular TransportAP

The plasma membrane is the border that makes a cell a cell, and in Unit 2 of the CED (Topics 2.4–2.9) the College Board expects you to treat it as a dynamic, selective structure — not a wall. The membrane is built from the phospholipids of Lesson 1: their amphipathic nature produces a bilayer with a hydrophobic core that small nonpolar molecules cross freely and almost everything else cannot. Everything a cell imports or exports is therefore regulated by membrane proteins, gradients and, when needed, ATP. If you master water potential calculations and the logic of tonicity in this lesson, you will collect points across the whole exam, because membrane questions recur in the plant, animal physiology and ecology units too.

The fluid mosaic model

The membrane is a fluid mosaic: phospholipids drift laterally like a two-dimensional liquid, while proteins float within or span the bilayer. Cholesterol, wedged between phospholipid tails in animal membranes, buffers fluidity — restraining movement at high temperatures and preventing tight packing at low temperatures. Integral proteins span the membrane and include channels, carriers and pumps; peripheral proteins sit on a surface. Carbohydrate chains on the outer surface (glycolipids and glycoproteins) act as identity markers for cell–cell recognition. Selective permeability follows directly from the chemistry: O₂, CO₂ and lipids diffuse through; ions and polar molecules such as glucose need a protein doorway.

Passive transport moves substances down their gradient and costs no ATP. Simple diffusion crosses the bilayer directly. Facilitated diffusion uses channel proteins (fast, often selective — aquaporins move water) or carrier proteins that change shape after binding. Osmosis is the diffusion of water across a selectively permeable membrane, always towards the side with higher solute concentration (lower free water).

Tonicity describes a solution relative to a cell. In a hypotonic solution (lower solute outside), water enters: animal cells swell and may lyse, while plant cells become turgid — their healthy state. In a hypertonic solution, water leaves: animal cells shrivel (crenation) and plant cytoplasm pulls away from the wall (plasmolysis). In an isotonic solution there is no net movement, the normal state for animal cells bathed in body fluids.

Water potential: the calculation the exam loves

Water moves from higher water potential (Ψ) to lower Ψ. The formula is Ψ = ΨP + ΨS, pressure potential plus solute potential, where ΨS = −iCRT: i is the ionisation constant (1 for sucrose, 2 for NaCl), C is molar concentration, R = 0.0831 L·bar/(mol·K), and T is temperature in kelvin (°C + 273). Solute potential is always zero or negative — adding solute can only lower Ψ. In an open container ΨP = 0, so Ψ = ΨS. A plant cell in pure water gains water until the wall pushes back; at equilibrium its positive ΨP exactly balances its negative ΨS and net movement stops, even though concentrations never equalise.

Active transport moves substances against their gradient using ATP. The sodium–potassium pump exports 3 Na⁺ and imports 2 K⁺ per ATP, building the electrochemical gradient that powers nerve impulses and cotransport: a carrier that lets Na⁺ rush back in can drag glucose uphill with it — secondary active transport. Large cargo uses vesicles: exocytosis fuses vesicles with the membrane to release contents; endocytosis buds the membrane inward (phagocytosis for solids, pinocytosis for fluids, receptor-mediated endocytosis for specific molecules such as cholesterol-carrying LDL).

Worked example 1: a full water potential problem

A plant cell has a solute potential of −0.8 bar and is placed in an open beaker of solution with ΨS = −0.3 bar. (a) State the initial water potential of the cell (assume ΨP = 0.1 bar). (b) Predict the direction of net water movement. (c) Describe the cell at equilibrium.

(a) Ψcell = ΨP + ΨS = 0.1 + (−0.8) = −0.7 bar. The beaker is open, so ΨP = 0 and Ψsolution = −0.3 bar. (b) Water moves from higher Ψ to lower Ψ: from −0.3 (solution) towards −0.7 (cell), so water enters the cell by osmosis. (c) As water enters, the protoplast presses on the wall and ΨP rises. Equilibrium is reached when Ψcell = Ψsolution = −0.3 bar, which needs ΨP = 0.5 bar. Note the trap the College Board sets: the cell does not reach equilibrium by its solute potential changing (its contents barely dilute) — pressure does the balancing.

Worked example 2: interpreting a dialysis-tubing osmosis experiment

Students filled dialysis tubing (permeable to water and small ions, not to sucrose) with a 20% sucrose solution and placed it in distilled water. The bag's mass rose from 10.0 g to 12.4 g in 30 minutes. A second bag filled with distilled water and placed in a 20% sucrose solution fell from 10.0 g to 8.1 g.

Calculate percentage change for each and explain the results. Bag 1: (12.4 − 10.0)/10.0 × 100 = +24%. Bag 2: (8.1 − 10.0)/10.0 × 100 = −19%. In bag 1 the inside is hypertonic to the surroundings, so water enters by osmosis; sucrose cannot leave through the tubing, only water crosses. In bag 2 the inside is hypotonic, so water leaves. Percentage change — not raw grams — is the correct comparison because the College Board designs these experiments with slightly different starting masses, and raw change would mislead. If a graph of percentage change against external sucrose concentration crosses zero, that x-intercept estimates the bag contents' own concentration — a favourite data-analysis question.

Where this shows up in the real world

Cystic fibrosis is a membrane-protein disease. The CFTR channel, which normally moves chloride ions (and, indirectly, water) across airway cell membranes, is misfolded and destroyed by the cell. Without chloride transport, airway mucus becomes dehydrated and thick, blocking airways and inviting infection. Haemodialysis applies the same principles externally: a patient's blood flows past a selectively permeable membrane so that urea diffuses out down its gradient while cells and proteins stay in — osmosis and diffusion doing medical work.

Common mistakes

• Reversing hypotonic and hypertonic. The prefix describes the solution's solute concentration, not the water flow. Hypotonic solution = less solute = water flows in.
• Forgetting ΨS is always ≤ 0. If you calculate a positive solute potential, you have dropped the negative sign from ΨS = −iCRT.
• Using Celsius in the formula. T must be in kelvin; a room-temperature value is about 298 K, not 25.
• Claiming equilibrium means equal concentrations. In a walled cell, equilibrium means equal water potentials, achieved by pressure.
• Calling facilitated diffusion active. Any protein-mediated movement down a gradient is passive, no matter how fancy the protein; ATP is the dividing line.
• Saying water moves from "high concentration of water to low" without tying it to solute. Acceptable, but the scoring point is the gradient of water potential / solute concentration.

AP Exam Experience

Question 1. A dialysis bag containing 0.4 M sucrose is placed in an open beaker of 0.1 M sucrose at 25 °C. The solute potential of the bag's contents is approximately (R = 0.0831 L·bar/mol·K)

• (A) −2.5 bar
• (B) −9.9 bar
• (C) +9.9 bar
• (D) −24.8 bar

Question 2. A freshwater fish constantly produces large volumes of very dilute urine and rarely drinks. The best explanation is that its cells are

• (A) hypertonic to the pond water, so water enters by osmosis and must be expelled
• (B) hypotonic to the pond water, so water leaves by osmosis
• (C) isotonic with the pond water, so kidneys conserve water
• (D) losing salts, so water follows the salts out of the body

Question 3. Glucose enters intestinal cells from the gut lumen even when glucose concentration inside the cell is already higher. This uptake depends on the sodium–potassium pump because

• (A) the pump transports glucose directly into the cell
• (B) the pump maintains the Na⁺ gradient that powers glucose cotransport into the cell
• (C) glucose is too large to pass through any membrane protein
• (D) the pump produces the glucose the cell absorbs

Answer key

• 1. (B) — ΨS = −iCRT = −(1)(0.4)(0.0831)(298) ≈ −9.9 bar. The beaker value (−2.5 bar) is the distractor for students who plug in the wrong concentration.
• 2. (A) — Fish body fluids have more solute than pond water, so water floods in across the gills; dilute urine is the disposal system.
• 3. (B) — This is secondary active transport: the pump burns ATP to keep intracellular Na⁺ low, and the downhill rush of Na⁺ through a cotransporter drags glucose uphill.

Free-response question. A student investigates osmosis using potato cores placed in sucrose solutions of different concentrations for 24 hours. The percentage change in mass is plotted against solution concentration, and the line crosses zero change at 0.35 M. (a) Explain what the zero-crossing concentration tells you about the potato tissue. (b) Describe the direction of net water movement when cores are placed in 0.1 M sucrose, and justify your answer. (c) The student repeats the experiment with the solutions at 5 °C instead of 22 °C. Predict how the rate of mass change will be affected and explain why. (d) Identify one variable that must be kept constant across treatments for the comparison to be valid, and explain why it matters.

Scoring guideline

• (a) 1 point: at 0.35 M there is no net movement, so the tissue's water potential equals the solution's — 0.35 M approximates the solute concentration of the potato cells.
• (b) 1 point: water enters the cores (mass increases). 1 point: justification — 0.1 M is hypotonic to the tissue / the tissue has lower water potential (more negative ΨS), so water moves in by osmosis.
• (c) 1 point: predicts a slower rate of mass change. 1 point: lower temperature means lower kinetic energy / slower molecular movement and diffusion.
• (d) 1 point: any valid controlled variable (same potato source, similar core size/surface area, same immersion time, same volume of solution) with a reason it affects mass change.