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Cells

Unit 2 of AP Biology, worth 10–13% of the exam. 13 questions below, each with the working. Every answer was checked by a second pass before it was published.

Cell structure and organelles, membrane structure and transport, osmosis and water potential, surface-area-to-volume, compartmentalisation, and the origins of cells.

How this unit is tested

Start by sorting facts into two buckets: what a structure IS (prokaryotic vs. eukaryotic, organelle identity) and what a structure DOES (function, especially how it interacts with membranes and gradients). Most AP questions test the second bucket — they give you a scenario (a cell in a solution, a molecule crossing a membrane, a cell growing in size) and ask you to predict a consequence, not just recall a name. For membrane and osmosis questions, always start by identifying water potential (Ψ) or solute concentration on each side, then remember water and solutes move down their gradients (high to low Ψ, or high to low concentration for simple diffusion) unless active transport or a pump is explicitly stated. Draw a quick sketch with two compartments and arrows if the question involves comparative concentrations — this catches sign errors before you commit to an answer. For surface-area-to-volume and compartmentalization questions, think in terms of ratios and rates: SA:V determines how fast a cell can exchange materials relative to how much it needs, and compartmentalization exists because it lets incompatible reactions (or reactions needing different pH/enzyme concentrations) happen in the same cell without interfering with each other. For origins-of-cells questions, treat the endosymbiotic theory as a hypothesis supported by specific structural evidence (not just 'mitochondria are useful') — the AP exam rewards citing the double membrane, own circular DNA, own ribosomes, and independent (binary-fission-like) division as the actual evidence.

What you have to know

Fluid Mosaic Model
The plasma membrane is a dynamic, fluid structure composed of a phospholipid bilayer in which proteins are embedded and can move laterally; membrane fluidity is affected by temperature, cholesterol content, and the saturation of fatty acid tails.
Water Potential Equation
Total water potential is $\Psi=\Psi_p+\Psi_s$, where $\Psi_p$ is pressure potential and $\Psi_s$ is solute potential. Solute potential is calculated as $\Psi_s=-iCRT$, where i is the ionization constant, C is molar concentration, R is the pressure constant (0.0831 L·MPa/(mol·K)), and T is temperature in Kelvin.
Direction of Water Movement
Water always moves across a selectively permeable membrane from a region of higher water potential to a region of lower water potential, regardless of which side has more solute.
Surface-Area-to-Volume Relationship
As a cell's linear dimension increases, volume increases with the cube of the dimension while surface area increases only with the square, so the surface-area-to-volume ratio decreases as cell size increases, limiting maximum cell size.
Endosymbiotic Theory
Mitochondria and chloroplasts are hypothesized to have originated as free-living prokaryotes that were engulfed by an ancestral eukaryotic cell and became permanent, mutually beneficial residents, evidenced by their double membranes, own circular DNA, own ribosomes, and independent division.

13 practice questions

  1. Explain why water moves from a hypotonic solution to a hypertonic solution across a selectively permeable membrane, in terms of water potential.
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    Answer. Because the hypertonic side has more dissolved solute, its solute potential (and thus overall water potential) is lower; water always moves from higher to lower water potential, so it flows from the hypotonic (higher Ψ) side to the hypertonic (lower Ψ) side.

    Adding solute lowers Ψs, which lowers total Ψ if pressure potential is unchanged. Water follows the water potential gradient, not the solute gradient directly, ending up moving toward the more concentrated solution.
  2. A red blood cell is placed in a 0.5 M NaCl solution, which is hypertonic relative to the cell's cytoplasm. Predict and explain what happens to the cell.
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    Answer. The cell will lose water by osmosis and shrink (crenate), because water moves from the cell's higher water potential to the solution's lower water potential.

    Since the surrounding solution has more solute, its water potential is lower than that of the cytoplasm. Water exits the cell down this gradient, causing the cell to shrivel.
  3. A plant cell is placed in distilled water. Describe what happens to the cell's pressure potential and volume as it reaches equilibrium.
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    Answer. Water enters the cell by osmosis, increasing cell volume and building up pressure potential (Ψp) against the rigid cell wall until Ψp rises enough to balance the negative solute potential, bringing total Ψ to equilibrium with the surrounding water (Ψ=0).

    Unlike an animal cell, a plant cell doesn't burst because the cell wall resists expansion, generating turgor pressure that opposes further net water entry.
  4. Calculate the solute potential of a 0.3 M glucose solution at 25°C (i = 1, R = 0.0831 L·MPa/(mol·K)). Show your work.
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    Answer. Ψs = -(1)(0.3)(0.0831)(298) ≈ -7.43 MPa

    Convert temperature to Kelvin (25+273=298), then plug into Ψs = -iCRT: 1 × 0.3 × 0.0831 × 298 ≈ 7.43, and the value is negative because solutes lower water potential.
  5. Compare passive transport and active transport in terms of energy requirement and direction of movement relative to the concentration gradient.
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    Answer. Passive transport (diffusion, facilitated diffusion, osmosis) requires no cellular energy and moves substances down their concentration gradient (high to low). Active transport requires ATP (or another energy source) and moves substances against their gradient (low to high).

    The key distinguishing feature is whether the cell must expend energy: passive processes exploit existing gradients, while active transport creates or maintains gradients that would otherwise dissipate.
  6. Explain the role of aquaporins in osmosis, and why root cells that absorb large amounts of water tend to have many of them.
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    Answer. Aquaporins are channel proteins that allow water to cross the membrane much faster than by diffusion through the lipid bilayer alone; root cells have many aquaporins because they need to move large volumes of water quickly to support the plant.

    Water can cross membranes on its own via simple diffusion, but this is slow; aquaporins provide dedicated channels that dramatically increase the rate of osmosis without requiring energy, since water still moves down its potential gradient.
  7. Which piece of evidence best supports the endosymbiotic theory of mitochondrial origin?
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    Answer. Mitochondria have a double membrane and their own circular DNA

    The double membrane (one from the original prokaryote, one from the host's engulfing vesicle), circular DNA, and independent 70S ribosomes are structural and genetic evidence consistent with an origin as a free-living prokaryote, unlike the other options which just describe mitochondrial function or location.
  8. Describe the pathway a protein destined for secretion takes through the endomembrane system, from synthesis to release.
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    Answer. The protein is synthesized by a ribosome on the rough ER, folded and modified inside the ER lumen, packaged into a transport vesicle, sent to the Golgi apparatus for further modification and sorting, packaged into a secretory vesicle, and released outside the cell by exocytosis when the vesicle fuses with the plasma membrane.

    This ordered pathway (rough ER → Golgi → secretory vesicle → exocytosis) illustrates how compartmentalization lets a protein be processed in stages, each in a distinct organelle with its own enzymatic environment.
  9. Explain why smaller cells have a higher surface-area-to-volume ratio than larger cells, and why this matters for nutrient uptake and waste removal.
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    Answer. Because volume scales with the cube of linear size while surface area scales with the square, smaller cells have proportionally more membrane surface per unit of volume; this higher SA:V ratio lets them exchange nutrients and wastes with their environment fast enough to meet the metabolic demands of their volume.

    As a cell grows, its metabolic needs (tied to volume) increase faster than its ability to exchange materials (tied to surface area), so past a certain size a cell cannot support itself and must either divide or adopt structures like microvilli.
  10. A spherical cell has a radius of 10 µm. If its radius doubles to 20 µm, by what factor do its surface area and volume each increase?
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    Answer. Surface area increases by a factor of 4; volume increases by a factor of 8.

    Surface area scales with r² (2²=4) and volume scales with r³ (2³=8) when radius doubles, so the SA:V ratio is cut in half, illustrating why doubling size makes exchange proportionally harder.
  11. Identify two structural features found in typical prokaryotic cells but absent in typical animal cells.
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    Answer. A cell wall (made of peptidoglycan) and a nucleoid region (unbound circular DNA) instead of a membrane-bound nucleus.

    Prokaryotes lack membrane-bound organelles entirely; their genetic material sits in an unenclosed nucleoid, and many also have a rigid cell wall for structural support, both of which distinguish them from animal cells.
  12. Explain how compartmentalization allows a eukaryotic cell to carry out chemical reactions that would be incompatible if they occurred in the same location.
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    Answer. By enclosing reactions inside separate membrane-bound organelles, a cell can maintain different local conditions (pH, enzyme concentration, substrate availability) needed for each process, such as acidic hydrolysis in lysosomes versus a near-neutral cytosol, preventing the reactions from interfering with or damaging each other.

    For example, lysosomal enzymes that digest macromolecules require a low pH; keeping them inside the lysosome protects the rest of the cell from being digested while still allowing the reaction to proceed.
  13. A solution has Ψs = -0.8 MPa and Ψp = 0.3 MPa. What is its total water potential, and if it is separated by a membrane from pure water (Ψ = 0 MPa), which direction will water move?
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    Answer. Total Ψ = 0.3 + (-0.8) = -0.5 MPa. Since the solution's water potential (-0.5 MPa) is lower than pure water's (0 MPa), water will move from the pure water into the solution.

    Adding Ψp and Ψs gives the net water potential; comparing it to the pure water side (Ψ=0) shows the solution is lower, so water flows down the gradient into the solution until equilibrium or a physical limit is reached.

What people get wrong

  1. Assuming water moves toward the hypotonic (low-solute) side because 'water follows water.' Instead, remember water moves toward the side with lower water potential, which is the side with MORE solute (hypertonic) when pressure potential is zero.
  2. Treating pressure potential (Ψp) as always positive or zero. In real plant systems, especially xylem, Ψp can be negative due to tension; don't drop it from the equation just because it 'should' be positive.
  3. Assuming facilitated diffusion requires ATP because it uses a protein channel or carrier. It is still passive transport — it moves solutes down their concentration gradient and needs no energy input, unlike active transport.
  4. Thinking bigger cells are more efficient because they have more total surface area. Compare the ratio, not the raw numbers: SA:V shrinks as cells grow, which is why cells stay small or fold their membranes (microvilli, cristae) instead of simply enlarging.
  5. Citing 'mitochondria make energy' as evidence for endosymbiosis. Function is not evidence of origin — the real evidence is structural and genetic: double membrane, own circular DNA, own 70S ribosomes, and division independent of the cell cycle.

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