Plasma Membranes — AP Biology Study Guide

For: AP Biology candidates sitting AP Biology.

Covers: The fluid mosaic model of membrane structure, selective permeability, passive and active transport, water potential calculations, osmosis, and experimental analysis of membrane function, aligned to AP Biology CED Unit 2 learning objectives.

You should already know: Basic structure of phospholipids, concentration gradients, cell theory.

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 Plasma Membranes?

The plasma membrane (also called the cell membrane or plasmalemma) is the selectively permeable phospholipid bilayer that forms the boundary of all living cells (prokaryotic and eukaryotic) and surrounds most membrane-bound organelles in eukaryotic cells. It is not merely a passive structural barrier: embedded proteins, carbohydrates, and lipid molecules carry out critical functions including transport of molecules, cell signaling, cell-cell recognition, and adhesion to extracellular matrices. According to the AP Biology Course and Exam Description (CED), Unit 2: Cell Structure and Function makes up 10-13% of the total AP exam score, and plasma membrane concepts account for roughly one-third of that unit, or 3-4% of your total exam score. Plasma membrane content appears in both multiple-choice questions (MCQ), where it often tests conceptual understanding of permeability and fluidity, and free-response questions (FRQ), where it frequently appears in data analysis of osmosis experiments or as a foundation for questions about cell signaling or homeostasis.

2. The Fluid Mosaic Model

The fluid mosaic model, first proposed by Singer and Nicolson in 1972, is the currently accepted model of plasma membrane structure. It describes the membrane as a "mosaic" of diverse embedded components (proteins, carbohydrates, lipids) embedded within a fluid phospholipid bilayer. Key components and properties include:

• Phospholipids: Amphipathic molecules with hydrophilic phosphate heads and hydrophobic fatty acid tails that spontaneously arrange into a bilayer in aqueous environments, with heads facing outward to the cytoplasm and extracellular fluid, and tails packed in the hydrophobic core.
• Fluidity: Phospholipids move laterally within their layer of the bilayer at a high rate, while rare "flip-flop" across layers is energetically unfavorable. Fluidity is regulated by temperature (lower temperatures reduce movement and packing increases), fatty acid saturation (unsaturated fatty acids have kinked tails that prevent packing, increasing fluidity), and cholesterol (a steroid lipid that buffers fluidity in animal cells: it reduces fluidity at high temperatures by restricting movement, and maintains fluidity at low temperatures by preventing tight packing of tails).
• Protein components: Integral proteins are embedded in the bilayer (often spanning the entire membrane as transmembrane proteins) with hydrophobic regions interacting with the core, while peripheral proteins are bound to the membrane surface (often to integral proteins) and do not enter the hydrophobic core.
• Carbohydrates: Bound to lipids (glycolipids) or proteins (glycoproteins) on the extracellular surface, used for cell-cell recognition and immune system function.

Worked Example

Problem: Two species of wildflower grow in different environments: Species 1 is adapted to the cold, 5°C average temperature of the Rocky Mountains alpine zone, and Species 2 is adapted to the 30°C average temperature of the Sonoran Desert. Predict the difference in fatty acid composition of the two species’ plasma membranes, and justify your prediction.

1. Lower temperatures reduce membrane fluidity by slowing phospholipid movement and allowing saturated fatty acid tails to pack tightly together, which can make the membrane too rigid to function.
2. To maintain fluidity at low temperatures, Species 1 (alpine) will have a higher proportion of unsaturated fatty acid tails than Species 2 (desert).
3. Unsaturated fatty acids have double bonds that create kinks in their tails, preventing tight packing even at low temperatures, which keeps the membrane fluid enough for protein movement and transport.
4. Species 2, adapted to high temperatures, will have a higher proportion of saturated fatty acid tails, which pack tightly to reduce excess fluidity and maintain membrane structural integrity at high temperatures.

Exam tip: Never mention cholesterol when answering questions about plant cell membrane fluidity—cholesterol is only used for fluidity regulation in animal cells. Plants exclusively adjust fatty acid saturation to modify fluidity, and AP exam graders will penalize incorrect references to cholesterol in plants.

3. Selective Permeability and Membrane Transport

The amphipathic structure of the phospholipid bilayer creates selective permeability, meaning only specific types of molecules can cross freely, while others require specialized transport proteins or energy input. Permeability is determined by two key properties of the solute: size and polarity/charge:

1. High permeability, no transport protein needed: Small nonpolar molecules (e.g., O₂, CO₂, steroid hormones) diffuse rapidly across the bilayer, as they are hydrophobic and interact favorably with the hydrophobic core.
2. Low permeability, very slow diffusion without proteins: Small polar uncharged molecules (e.g., water, urea) can diffuse slowly through the bilayer, but most movement occurs through specialized channel proteins called aquaporins.
3. No permeability without transport proteins: Large polar molecules (e.g., glucose, sucrose) and all charged ions (e.g., Na⁺, K⁺, Cl⁻) cannot cross the hydrophobic core at all, and require channel or carrier proteins for transport.

Membrane transport is divided into two broad categories: passive transport, which does not require energy input and moves solutes down their concentration gradient (includes simple diffusion and facilitated diffusion), and active transport, which requires energy input (ATP or an existing electrochemical gradient) to move solutes against their concentration gradient. Bulk transport (endocytosis and exocytosis) is a form of active transport that moves large particles or volumes of fluid across the membrane via vesicles.

Worked Example

Problem: Which of the following molecules will diffuse most rapidly across a pure phospholipid bilayer without any transport proteins? A) Glucose, B) Potassium ion, C) Carbon dioxide, D) Sucrose

1. First, eliminate solutes that cannot cross the bilayer at all without transport proteins: Potassium (B) is a charged ion, which cannot interact with the hydrophobic core, so eliminate B.
2. Next, eliminate large polar molecules: Glucose (A) and sucrose (D) are both large polar molecules that are too big and polar to diffuse through the bilayer core, so eliminate A and D.
3. Carbon dioxide (C) is a small nonpolar molecule, which is hydrophobic and interacts favorably with the bilayer core, so it diffuses rapidly across the membrane without assistance.
4. The correct answer is C.

Exam tip: If an AP question specifies diffusion "without any transport proteins", remember that even small polar molecules like water diffuse much more slowly than small nonpolar gases, so the nonpolar molecule will almost always be the correct answer.

4. Water Potential and Osmosis

Osmosis is the net diffusion of free water across a selectively permeable membrane. To predict the direction of osmosis, we use the concept of water potential (symbol Ψ, the Greek letter psi), which measures the potential energy of water to move. Water always moves from an area of higher water potential (less negative) to an area of lower water potential (more negative). The total water potential of a solution is calculated as: $$\Psi ={\Psi }_s+{\Psi }_p$$ Where ${\Psi }_s$ = solute potential (osmotic potential), and ${\Psi }_p$ = pressure potential. Solute potential is calculated using: $${\Psi }_s=-iCRT$$ Where:

• $i$ = ionization constant: the number of ions a solute dissociates into in water (1 for covalent solutes like sucrose, 2 for NaCl, etc.)
• $C$ = molar concentration of the solute (mol/L)
• $R$ = pressure constant, $0.0831\frac{L\cdot bar}{mol\cdot K}$ for AP Biology calculations
• $T$ = temperature in Kelvin (°C + 273)

Solute potential is always negative, because adding solute reduces the number of free water molecules, lowering water potential. Pressure potential is 0 for open systems (like an open beaker, or soil at atmospheric pressure), positive in turgid plant cells (the cell wall pushes back on the cytoplasm, creating positive pressure), and negative in plant xylem when water is being pulled up to the leaves.

Worked Example

Problem: Calculate the total water potential of a 0.4M sucrose solution in an open beaker at 27°C. Show your work.

1. First, list all known values: Sucrose is a covalent solute, so $i=1$. $C=0.4\frac{mol}{L}$. $R=0.0831\frac{L\cdot bar}{mol\cdot K}$. Convert temperature to Kelvin: $27+273=300K$.
2. The beaker is open to the atmosphere, so pressure potential ${\Psi }_p=0$ bar.
3. Calculate solute potential: ${\Psi }_s=-(1)(0.4)(0.0831)(300)=-9.97\approx -10.0$ bar.
4. Calculate total water potential: $\Psi ={\Psi }_s+{\Psi }_p=-10.0+0=-10.0$ bar.

Exam tip: Always double-check that you converted temperature to Kelvin before plugging it into the solute potential formula. Using Celsius instead of Kelvin is one of the most common mistakes AP graders see on water potential FRQs, and it will cost you points.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Claiming cholesterol regulates membrane fluidity in plant cells. Why: Students memorize cholesterol's role in animal cells and incorrectly generalize it to all cell types. Correct move: Always note that only animal cells use cholesterol for fluidity buffering; plant cells adjust fatty acid saturation to change fluidity.
• Wrong move: Confusing water potential direction with solute concentration. Why: Students remember water moves from low solute to high solute, but mix up the direction when working with negative water potential values. Correct move: Always use the rule "water moves from higher (less negative) Ψ to lower (more negative) Ψ" to double-check direction regardless of solute concentration.
• Wrong move: Assuming all polar molecules cannot cross the plasma membrane. Why: Students overgeneralize the hydrophobic core rule to all polar molecules, regardless of size. Correct move: Explicitly separate small polar molecules (water, urea) that can diffuse slowly across the bilayer from large polar molecules (glucose, sucrose) that require transport proteins.
• Wrong move: Using $i=1$ for all solutes in water potential calculations. Why: Students forget that ionic solutes dissociate into multiple ions, which changes the total number of solute particles. Correct move: First identify if the solute is ionic (i>1) or covalent (i=1) before plugging values into the formula.
• Wrong move: Claiming all molecules moved in active transport go against their concentration gradient. Why: Students forget that secondary active transport uses the gradient of one molecule to power movement of another against its gradient. Correct move: Remember active transport is defined by requiring energy input, not the direction of all solutes being transported.
• Wrong move: Classifying peripheral proteins as any protein on the membrane surface and all integral proteins as transmembrane. Why: Students overgeneralize common diagrams that show all integral proteins spanning the bilayer. Correct move: Classify proteins as integral if they have at least one hydrophobic region embedded in the bilayer core, regardless of whether they span the entire membrane.

6. Practice Questions (AP Biology Style)

Question 1 (Multiple Choice)

A researcher places a red blood cell with an internal solute concentration of 0.15M into an open beaker containing a 0.3M sucrose solution at 20°C. Red blood cells have no cell wall, and pressure potential inside the cell is approximately 0 bar. Which of the following describes what will happen to the cell? A) The cell will take up water and eventually burst B) The cell will lose water and shrivel C) No net movement of water will occur D) Water will be pumped into the cell via active transport

Worked Solution: First calculate the water potential of the cell and the beaker solution. For the red blood cell: i=1 for total solutes, Ψ_s = -(1)(0.15)(0.0831)(293) ≈ -3.65 bar, Ψ_p=0, so Ψ_cell ≈ -3.65 bar. For the beaker solution: Ψ_s = -(1)(0.3)(0.0831)(293) ≈ -7.3 bar, Ψ_p=0, so Ψ_beaker ≈ -7.3 bar. Water moves from higher water potential (-3.65 bar, cell) to lower water potential (-7.3 bar, beaker), so net movement of water is out of the cell. The beaker solution is hypertonic to the cell, so the cell loses water and shrivels. Correct answer: B.

Question 2 (Free Response)

Membrane fluidity is a critical property that allows cells to adapt to changing environmental temperatures. (a) Describe the amphipathic nature of phospholipids, and explain why this property leads to spontaneous bilayer formation in aqueous environments. (b) A freshwater fish moves from a 25°C aquarium to a 5°C cold stream. Predict how the phospholipid composition of the fish's cell membranes will change over 48 hours in the cold stream, and justify your prediction. (c) Many enzymes embedded in the plasma membrane require lateral movement within the bilayer to interact with their substrates. Explain how the change you predicted in (b) will maintain enzyme function at low temperatures.

Worked Solution: (a) Phospholipids are amphipathic because they have a polar, hydrophilic phosphate head group and two nonpolar, hydrophobic fatty acid tails. In aqueous solution, the hydrophilic heads spontaneously interact with polar water molecules, while the hydrophobic tails cluster together away from water to minimize unfavorable interactions. This arrangement forms a stable bilayer with heads on the outer and inner aqueous surfaces and tails in the hydrophobic core. (b) Prediction: The proportion of unsaturated fatty acids in the membrane phospholipids will increase. Justification: Lower temperatures reduce membrane fluidity by slowing phospholipid movement and allowing saturated fatty acid tails to pack tightly together. Unsaturated fatty acids have double bonds that create kinks in their tails, which prevent tight packing and maintain membrane fluidity at lower temperatures. (c) The membrane-bound enzymes rely on fluidity to diffuse laterally within the bilayer and encounter their substrates. Increasing unsaturated fatty acid content maintains sufficient fluidity at 5°C, allowing enzymes to move and interact with substrates, which preserves enzyme activity and cellular function. If fluidity was too low, enzyme movement would stop, and cellular reactions would slow or halt.

Question 3 (Application / Real-World Style)

Crop growers in coastal regions often deal with saline soil (high salt concentration) from ocean flooding. A plant breeder tests two new varieties of cotton for salt tolerance: Variety X and Variety Y. When grown in soil with a 0.15M NaCl solution (open to atmospheric pressure) at 25°C, Variety X root cells have a total internal solute concentration of 0.4M solute particles, and a turgor pressure potential of +2 bar. Variety Y root cells have a total internal solute concentration of 0.2M solute particles, and the same turgor pressure potential of +2 bar. Using water potential calculations, predict which variety will survive in the saline soil, and explain your result.

Worked Solution: First calculate the water potential of the saline soil solution: NaCl dissociates into 2 ions, so i=2, C=0.15M, T=25+273=298K, Ψ_p=0 for open soil. Ψ_soil = (-iCRT) + 0 = -(2)(0.15)(0.0831)(298) ≈ -7.43 bar. For Variety X: Ψ_s = -(0.4)(0.0831)(298) ≈ -9.90 bar. Ψ_X = -9.90 + 2 = -7.90 bar. For Variety Y: Ψ_s = -(0.2)(0.0831)(298) ≈ -4.95 bar. Ψ_Y = -4.95 + 2 = -2.95 bar. Water moves from higher Ψ to lower Ψ. For roots to take up water from soil, root Ψ must be lower (more negative) than soil Ψ. Variety X has Ψ=-7.90 bar < -7.43 bar (soil Ψ), so water moves into Variety X roots, allowing it to grow. Variety Y has Ψ=-2.95 bar > -7.43 bar, so water moves out of Variety Y roots, causing the plant to wilt and die. Variety X is salt-tolerant and will survive, because it accumulates enough solutes to lower its water potential below that of the saline soil, enabling water uptake.

7. Quick Reference Cheatsheet

8. What's Next

Plasma membrane structure and function is the foundational prerequisite for all topics related to cellular transport, cell signaling, and homeostasis across the AP Biology course. Immediately next, you will apply your understanding of selective permeability and water potential to the movement of water and solutes in plant vascular systems, and then to the mechanism of cell signaling via membrane-bound receptors. Without mastering the core concepts of plasma membranes here, you will not be able to correctly interpret data from the AP Biology osmosis lab (a common exam topic) or explain how signal transduction pathways initiate at the cell surface. This topic also feeds into the overarching theme of how cells maintain homeostasis, which is tested across every unit of the exam.

• Membrane Transport
• Water Potential
• Cell Signaling
• Osmoregulation