Cell Structure and Function — AP Biology Study Guide

For: AP Biology candidates sitting AP Biology.

Covers: Surface area-to-volume ratio calculations, compartmentalization in prokaryotes vs eukaryotes, fluid mosaic model of cell membranes, membrane permeability, tonicity and osmolarity, and the endosymbiotic theory, aligned with AP Biology CED learning objectives.

You should already know: Basic classification of prokaryotic vs eukaryotic cells, amphipathic structure of phospholipids, general definitions of passive and active transport.

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 Cell Structure and Function?

Cell Structure and Function is the core study of how the physical organization of cellular components directly enables their biological roles, connecting micro-scale structure to whole-cell and organismal function. Aligned with the AP Biology Course and Exam Description (CED), this topic accounts for 10–13% of total exam score weight, and questions appear on both multiple-choice (MCQ) and free-response (FRQ) sections. Unlike introductory memorization of organelle names, the AP exam focuses on predicting how changes in structure alter function, rather than just recalling what each organelle does. This topic is the foundational prerequisite for nearly all subsequent units in AP Biology, from membrane transport to cell respiration to cell division, as all biological processes at the cellular level depend on the structure of the components that carry them out.

2. Surface Area-to-Volume Ratio

All cells rely on diffusion across their outer cell membrane to obtain nutrients and eliminate waste. The rate of diffusion depends on the amount of membrane surface area available for exchange, while the cell’s demand for nutrients and production of waste depends on its volume. For any cell of a given shape, as cell size increases, volume increases proportionally faster than surface area. For a cuboidal cell, the surface area-to-volume ratio (SA/V) simplifies to: $$\frac{SA}{V}=\frac{6s^2}{s^3}=\frac{6}{s}$$ where $s$ = side length of the cube. A higher SA/V ratio means more surface area per unit of cytoplasmic volume, which supports faster, more efficient exchange of materials. This is why cells remain small: large cells cannot exchange materials fast enough to support their metabolic needs. Specialized cells that require high exchange rates (e.g., intestinal epithelial cells, root hair cells) evolve folded membrane structures that increase surface area without significantly increasing volume, thus boosting SA/V ratio.

Worked Example

A student compares three spherical yeast cells: Cell 1 has radius 2 μm, Cell 2 has radius 4 μm, Cell 3 has radius 8 μm. For a sphere, $SA=4\pi r^2$ and $V=\frac{4}{3}\pi r^3$. Which cell is most efficient at glucose uptake, and what is its SA/V ratio?

1. Simplify the SA/V ratio for a sphere: $\frac{4\pi r^2}{\frac{4}{3}\pi r^3}=\frac{3}{r}$
2. Calculate the ratio for each cell: Cell 1 = $3/2=1.5\mu m^{-1}$, Cell 2 = $3/4=0.75\mu m^{-1}$, Cell 3 = $3/8=0.375\mu m^{-1}$
3. Higher SA/V ratio means more surface area per unit volume for glucose transport across the membrane, leading to more efficient uptake.
4. Cell 1 (radius 2 μm) has the highest SA/V ratio and is most efficient.

Exam tip: On FRQs, always connect a folded membrane structure to SA/V ratio and link it explicitly to the cell’s function (e.g., "microvilli increase SA/V to increase nutrient absorption rate") — generic statements about "more surface area" will not earn full points.

3. Fluid Mosaic Model and Membrane Permeability

The fluid mosaic model describes the structure of the cell membrane as a dynamic bilayer of amphipathic phospholipids embedded with proteins, cholesterol, and carbohydrate chains. Phospholipids have hydrophilic phosphate heads that face the aqueous extracellular and intracellular environments, and hydrophobic fatty acid tails that face the interior of the bilayer. The membrane is "fluid" because phospholipids can move laterally within the bilayer, and "mosaic" because it is made of many different embedded components.

Membrane permeability depends on the properties of the solute and the hydrophobic core of the bilayer: small nonpolar molecules (e.g., O₂, CO₂, steroid hormones) can dissolve in the hydrophobic core and cross freely. Small uncharged polar molecules (e.g., water) can cross slowly, but most cross via specialized channel proteins called aquaporins. Large polar molecules (e.g., glucose, sucrose) and all charged molecules/ions (e.g., Na⁺, Cl⁻, amino acids) cannot cross the hydrophobic core spontaneously and require transport proteins to cross the membrane. Fluidity is regulated by fatty acid saturation and cholesterol: unsaturated fatty acids have kinked tails that prevent packing, increasing fluidity at low temperatures, while cholesterol has a dual role: it reduces fluidity at high temperatures by restricting movement, and increases fluidity at low temperatures by preventing tight packing.

Worked Example

A synthetic phospholipid bilayer (no embedded proteins) is placed between two chambers: Chamber A contains 1 mM glucose, 1 mM O₂, 1 mM testosterone, and 1 mM Ca²⁺. Predict which solutes will move to Chamber B by diffusion, and justify your answer.

1. Categorize each solute by size and polarity: O₂ is small nonpolar, testosterone is a small nonpolar steroid, glucose is large polar, Ca²⁺ is a charged ion.
2. The hydrophobic core of the bilayer only allows nonpolar and small uncharged solutes to cross; it blocks charged and large polar solutes due to hydrophobic exclusion of polar/charged substances.
3. Since the bilayer has no transport proteins to facilitate movement of glucose or Ca²⁺, these solutes cannot cross.
4. Only O₂ and testosterone will diffuse spontaneously from Chamber A to Chamber B.

Exam tip: Never generalize that all polar molecules cannot cross the membrane; always specify that large polar and all charged molecules cannot cross, while small uncharged polar molecules can cross slowly without transport proteins.

4. Tonicity and Osmolarity

Osmosis is the net diffusion of water across a selectively permeable membrane, from a region of lower non-penetrating solute concentration to higher non-penetrating solute concentration. Tonicity describes the ability of an extracellular solution to cause net water movement into or out of a cell, which depends entirely on the concentration of non-penetrating solutes (solutes that cannot cross the membrane). A solution is hypertonic if it has a higher concentration of non-penetrating solutes than the cell (net water out of the cell), hypotonic if it has a lower concentration (net water into the cell), and isotonic if concentrations are equal (no net water movement).

A common lab calculation on the AP exam is percent change in mass of a cell/tissue placed in a solution, calculated as: $$\%\Delta m=\frac{m_{final}-m_{initial}}{m_{initial}}\times 100\%$$ Positive percent change means the cell gained mass (water moved in, solution is hypotonic), negative means the cell lost mass (water moved out, solution is hypertonic).

Worked Example

A carrot cube with an initial mass of 4.8 g is placed in a 0.3 M sucrose solution (sucrose cannot cross the carrot cell membrane). After 1 hour, the cube has a final mass of 5.2 g. Calculate the percent change in mass, and identify the tonicity of the solution relative to the carrot cell cytoplasm.

1. Write the percent change formula: $\%\Delta m=\frac{m_f-m_i}{m_i}\times 100\%$
2. Substitute values: $\%\Delta m=\frac{5.2-4.8}{4.8}\times 100\%=\frac{0.4}{4.8}\times 100\%\approx +8.3\%$
3. A positive percent change means the carrot cube gained mass, so net water moved into the cell from the extracellular solution.
4. Net water movement into the cell occurs when the extracellular solution has a lower concentration of non-penetrating solutes, so the solution is hypotonic relative to the carrot cytoplasm.

Exam tip: If the membrane is permeable to the solute, the solute will diffuse across the membrane to equalize concentration, so even if starting solute concentrations differ, there will be no net osmosis after equilibration. Always check if the solute is penetrating first.

5. Compartmentalization and Endosymbiotic Theory

Unlike prokaryotes, eukaryotic cells have membrane-bound organelles that compartmentalize specific cellular functions. Compartmentalization provides two key advantages: (1) it separates incompatible chemical reactions (e.g., keeping digestive enzymes in lysosomes away from the cytoplasm), and (2) it concentrates enzymes and substrates for specific reactions, increasing reaction rate, and creates more surface area for membrane-bound processes like ATP synthesis.

The endosymbiotic theory states that mitochondria and chloroplasts evolved from free-living prokaryotes that were engulfed by a ancestral eukaryotic cell, forming a symbiotic relationship. Key evidence for this theory includes: (1) mitochondria and chloroplasts have double membranes (consistent with engulfment), (2) they have circular DNA similar to prokaryotic chromosomes, (3) they have ribosomes that are similar in size and structure to prokaryotic ribosomes, (4) they replicate independently of the host cell via binary fission, like prokaryotes.

Worked Example

The inner membrane of mitochondria is highly folded into cristae, while the outer membrane is smooth. Explain how this structure supports the function of mitochondria in ATP production.

1. The folding of the inner membrane greatly increases its surface area without increasing the volume of the organelle.
2. The enzymes and electron transport chain proteins required for oxidative ATP production are embedded in the inner mitochondrial membrane.
3. More surface area means more space for electron transport chain proteins and ATP synthase enzymes, which increases the total amount of ATP the mitochondrion can produce.
4. Compartmentalization also allows the mitochondrion to maintain a proton gradient across the inner membrane, which is required for ATP synthesis via chemiosmosis.

Exam tip: When listing evidence for endosymbiotic theory, avoid vague statements like "they have their own DNA"; specify that it is circular DNA (unlike linear eukaryotic nuclear DNA) to earn full points.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Claiming that large cells of the same shape have a higher surface area-to-volume ratio than small cells. Why: Students confuse total surface area with surface area per unit volume; large cells have more total SA but lower SA/V. Correct move: Always remember that for cells of the same shape, SA/V decreases as cell size increases, and explicitly state this on exam questions.
• Wrong move: Calling a solution hypertonic just because it has a higher starting solute concentration, even when the solute can freely cross the membrane. Why: Students forget that tonicity only depends on non-penetrating solutes that drive net water movement after equilibration. Correct move: First confirm whether the solute is penetrating or non-penetrating before assigning tonicity.
• Wrong move: Claiming that increasing cholesterol in the membrane always decreases fluidity, regardless of temperature. Why: Students memorize one effect of cholesterol and forget its dual temperature-dependent role. Correct move: State that cholesterol reduces fluidity at high temperatures and increases fluidity at low temperatures to maintain homeostasis.
• Wrong move: Stating that prokaryotes have no membrane structures, only eukaryotes do. Why: Students confuse membrane-bound organelles with all cellular membranes. Correct move: State that prokaryotes lack membrane-bound organelles (e.g., nucleus, mitochondria), not any membrane structures at all.
• Wrong move: Arguing that compartmentalization is only useful for separating harmful enzymes from the cytoplasm. Why: Students only remember one function of compartmentalization and miss the other key advantage. Correct move: Always mention both incompatible reaction separation and increased reaction rate via concentrated enzymes/substrates when justifying the benefit of compartmentalization.

7. Practice Questions (AP Biology Style)

Question 1 (Multiple Choice)

Algal cells adapted to grow at 5°C in Arctic lakes are compared to algal cells of the same species adapted to grow at 25°C in temperate lakes. Which of the following combinations of characteristics would you expect to see in the Arctic-adapted cells, compared to the temperate-adapted cells? A) Higher proportion of saturated fatty acids, higher cholesterol content B) Higher proportion of unsaturated fatty acids, higher cholesterol content C) Higher proportion of saturated fatty acids, lower cholesterol content D) Higher proportion of unsaturated fatty acids, lower cholesterol content

Worked Solution: Low temperatures cause phospholipids to pack tightly together, reducing membrane fluidity. Unsaturated fatty acids have kinked tails that prevent tight packing, so Arctic-adapted cells need a higher proportion of unsaturated fatty acids to maintain fluidity. Cholesterol also increases membrane fluidity at low temperatures by preventing tight packing of phospholipids, so Arctic cells will have higher cholesterol content to stabilize fluidity. This matches option B. The correct answer is B.

Question 2 (Free Response)

A researcher is investigating the origin of eukaryotic cells and isolates a new organelle from a species of amoeba. She collects the following observations: the organelle has two membranes, its DNA is linear, and it has ribosomes that match the size of eukaryotic cytoplasmic ribosomes. (a) Based on these data, is this organelle likely a product of endosymbiosis? Justify your answer. (b) Explain two advantages of compartmentalization in eukaryotic cells that prokaryotes cannot achieve. (c) Prokaryotic cells are generally smaller than eukaryotic cells. Explain how cell size affects the ability of prokaryotes to carry out metabolic processes.

Worked Solution: (a) No, this organelle is not likely a product of endosymbiosis. Key evidence for endosymbiosis is that mitochondria and chloroplasts have circular DNA (like prokaryotes) and prokaryotic-sized ribosomes. This organelle has linear DNA and eukaryotic-sized ribosomes, which does not match the predicted pattern for an endosymbiotic organelle. (b) First, compartmentalization allows eukaryotes to separate incompatible chemical reactions: for example, digestive enzymes in lysosomes are kept separate from the cytoplasm, preventing damage to cellular macromolecules. Second, compartmentalization concentrates enzymes and substrates for specific reactions, increasing reaction rate, and allows for the formation of proton gradients across organelle membranes that are required for ATP synthesis. (c) Smaller prokaryotic cells have a higher surface area-to-volume ratio than larger eukaryotic cells. This allows prokaryotes to exchange nutrients and waste efficiently across their cell membrane, supporting fast metabolic rates without the need for internal membrane-bound organelles to produce ATP.

Question 3 (Application / Real-World Style)

A student conducts a beetroot osmosis experiment to find the osmolarity of beetroot root cells. She cuts 5 identical 5g beetroot cubes and places each in a different NaCl solution (NaCl is non-penetrating). After 2 hours, she records the final mass of each cube: 5.6g (0.0 M), 5.3g (0.1 M), 5.0g (0.2 M), 4.7g (0.3 M), 4.4g (0.4 M). What is the osmolarity of the beetroot cells, and what does this mean in context?

Worked Solution: Osmolarity of the beetroot cells is equal to the NaCl concentration where there is no net change in mass, which means the solution is isotonic to the cells. Calculate percent change for each solution: 0.0 M = +12%, 0.1 M = +6%, 0.2 M = 0%, 0.3 M = -6%, 0.4 M = -12%. There is no net change in mass at 0.2 M NaCl. The osmolarity of non-penetrating solutes in the beetroot root cells is approximately 0.2 osmol/L. In context, when beetroot cells are placed in a 0.2 M NaCl solution, no net osmosis occurs because the solute concentration inside and outside the cells is equal.

8. Quick Reference Cheatsheet

9. What's Next

This chapter is the foundational prerequisite for all subsequent topics that depend on understanding how cellular structure drives function. Immediately next, you will study membrane transport and cell signaling, which rely entirely on the fluid mosaic model and principles of permeability and compartmentalization you learned here. Without mastering how membrane structure impacts protein function and solute movement, you will not be able to explain signal transduction pathways or transport disorders, which are heavily tested on the AP exam. This topic also feeds into bigger concepts across the course: cell respiration and photosynthesis depend on the structure of mitochondria and chloroplasts, which is explained by endosymbiotic theory, and cell division relies on the SA/V relationship to explain why cells divide rather than growing indefinitely.

• Membrane Transport
• Cell Communication and Signal Transduction
• Cellular Respiration
• Cell Cycle and Cell Division