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AP Biology Unit 2 Practice Questions: Cell Structure and Function

10 original exam-style questions on Cell Structure and Function. Answer each one to see the explanation — no account needed.

Question 1 of 10 · Enzyme Structure and Function

A student investigates the effect of pH on the activity of salivary amylase, an enzyme that hydrolyzes starch into maltose. She prepares identical starch-buffer solutions at varying pH values and adds equal amounts of enzyme extract. After 10 minutes at 37°C, she measures remaining starch concentration using iodine absorbance and calculates relative enzyme activity.

Table 1. Effect of pH on Salivary Amylase Activity
pHRemaining Starch (mg/mL)Relative Activity (%)
3.09.82
5.07.129
6.51.288
7.00.9100
8.02.476
9.06.337
11.09.64
A researcher hypothesizes that replacing the histidine residues in the enzyme's active site with alanine (which has an uncharged, non-ionizable side chain) would shift the optimal pH. Which of the following predictions is best supported by the data and the hypothesis?
  1. A. The mutant enzyme would show maximum activity at the same pH of 7.0, because the overall protein fold is unchanged
  2. B. The mutant enzyme would lose all catalytic activity at every pH tested, because alanine cannot participate in any catalysis
  3. C. The mutant enzyme's optimal pH could shift, because histidine's ionization (pKa ≈ 6.0) contributes to the active site's charge environment at physiological pH
  4. D. The mutant enzyme would show maximum activity at a lower pH, because alanine destabilizes the substrate binding pocket at neutral pH
Show answer and explanation

Correct answer: C

Histidine has a pKa near 6.0, meaning it switches between protonated and deprotonated forms close to physiological pH; its ionization state directly influences the active site's electrostatic environment and thus the optimal pH. Replacing it with non-ionizable alanine would eliminate this pH-sensitive contribution and could shift the optimum. Choice A is incorrect because removing an ionizable residue that contributes to catalysis will alter the pH–activity profile even if the global protein fold is largely preserved.

Question 2 of 10 · Osmosis and Water Potential

A researcher investigates osmosis in plant cells by placing leaf discs in solutions of varying sucrose concentrations and measuring the change in mass after 30 minutes.

Mass Change of Leaf Discs in Sucrose Solutions
Sucrose Concentration (M)Initial Mass (g)Final Mass (g)% Change in Mass
0.02.002.24+12
0.22.002.12+6
0.42.002.01+0.5
0.62.001.88-6
0.82.001.74-13
Which of the following experimental changes would most likely shift the isotonic point to a lower sucrose concentration?
  1. A. Using a plant species with a lower cytoplasmic solute concentration
  2. B. Using a plant species with a higher cytoplasmic solute concentration
  3. C. Increasing the temperature of the sucrose solutions
  4. D. Replacing sucrose with glucose solutions of the same concentration
Show answer and explanation

Correct answer: A

The isotonic point corresponds to the external solute concentration equal to the cell's internal solute concentration; if cells have a lower internal solute concentration, equilibrium (no mass change) would occur at a lower external sucrose concentration. Choice B would shift the isotonic point higher, not lower.

Question 3 of 10 · Endosymbiotic Theory

The theory of endosymbiosis proposes that mitochondria and chloroplasts evolved from free-living prokaryotes engulfed by a host cell. Which of the following pieces of evidence most strongly supports this theory?
  1. A. Mitochondria and chloroplasts are surrounded by a single membrane, just like prokaryotes
  2. B. Mitochondria and chloroplasts replicate by meiosis, like eukaryotic organelles
  3. C. Mitochondria and chloroplasts can be stained with the same dyes used for the nucleus
  4. D. Mitochondria and chloroplasts contain their own circular DNA and 70S ribosomes, similar to those of bacteria
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Correct answer: D

The presence of circular DNA (rather than linear chromosomes) and 70S ribosomes (the prokaryotic type, not the 80S eukaryotic ribosomes) in mitochondria and chloroplasts is strong molecular evidence for their prokaryotic ancestry. Choice A is incorrect because both organelles are bounded by double membranes, not a single membrane.

Question 4 of 10 · Endomembrane System

A protein destined for secretion is synthesized by ribosomes and must travel through the endomembrane system before being released from the cell. Which of the following sequences correctly describes the route this protein takes?
  1. A. Free ribosome → smooth ER → Golgi apparatus → lysosome → extracellular space
  2. B. Rough ER → mitochondria → Golgi apparatus → plasma membrane
  3. C. Rough ER → Golgi apparatus → secretory vesicle → plasma membrane → extracellular space
  4. D. Nuclear envelope → rough ER → secretory vesicle → extracellular space
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Correct answer: C

Secretory proteins are synthesized on rough ER-bound ribosomes, enter the ER lumen, are packaged into vesicles for transport to the Golgi apparatus for processing, then sorted into secretory vesicles that fuse with the plasma membrane to release their contents by exocytosis. Free ribosomes (A) produce cytosolic proteins destined for the cytoplasm, not secretion.

Question 5 of 10 · Facilitated Diffusion and Transport Proteins

A research team investigates membrane permeability by testing the rate at which various molecules cross an artificial phospholipid bilayer (containing no transport proteins). The results are recorded below.

Permeability of Artificial Lipid Bilayer to Various Molecules
MoleculeSizePolarity/ChargeRelative Permeability
O2O_2SmallNonpolarVery High
CO2CO_2SmallNonpolarVery High
H2OH_2OSmallPolarModerate
GlucoseLargePolarVery Low
Na+Na^+Small ionChargedNegligible
Amino acidLargePolarNegligible
In a living cell, glucose crosses the plasma membrane much more rapidly than predicted by these data. Which of the following best accounts for this difference?
  1. A. Living cells have transport proteins (e.g., GLUT transporters) that facilitate glucose movement across the membrane
  2. B. Glucose is converted to a nonpolar form before crossing the membrane in living cells
  3. C. Living cells heat the membrane, increasing glucose diffusion rate through the bilayer
  4. D. The concentration gradient of glucose is steeper in living cells, increasing the simple diffusion rate
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Correct answer: A

Living cells express specific carrier proteins (GLUT transporters) that enable facilitated diffusion of glucose across the membrane. The artificial bilayer in the experiment lacks these proteins, explaining the very low permeability for glucose in the model versus a living cell.

Question 6 of 10 · Endomembrane System

Which of the following correctly distinguishes the endomembrane system from the mitochondria in a eukaryotic cell?
  1. A. The endomembrane system generates ATP, while mitochondria synthesize proteins for secretion
  2. B. The endomembrane system is found only in plant cells, while mitochondria are found only in animal cells
  3. C. The endomembrane system and mitochondria both evolved from endosymbiotic bacteria
  4. D. The endomembrane system (ER, Golgi, vesicles) processes and routes proteins and lipids, while mitochondria produce ATP through cellular respiration
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Correct answer: D

The endomembrane system (rough ER, smooth ER, Golgi apparatus, vesicles) is a membrane-trafficking network that synthesizes, modifies, and directs proteins and lipids. Mitochondria are energy-converting organelles that produce ATP via oxidative phosphorylation. Only mitochondria (and chloroplasts) are thought to have endosymbiotic origins (C is partially correct but wrongly includes the endomembrane system).

Question 7 of 10 · Active Transport

The sodium-potassium pump (Na+/K+Na^+/K^+-ATPase) moves 3 Na+Na^+ out of the cell and 2 K+K^+ into the cell for each ATP hydrolyzed. Which of the following best describes this process?
  1. A. Active transport, because both ions are moved against their electrochemical gradients using ATP energy
  2. B. Passive transport, because both ions move down their concentration gradients
  3. C. Facilitated diffusion, because the pump uses a protein channel to move ions
  4. D. Osmosis, because charged particles are crossing the selectively permeable membrane
Show answer and explanation

Correct answer: A

The Na+/K+Na^+/K^+ pump uses ATP to move Na+Na^+ out and K+K^+ in, both against their respective electrochemical gradients, which is the defining characteristic of primary active transport. Choice C is incorrect because facilitated diffusion requires no energy input and moves solutes down their gradient.

Question 8 of 10 · Membrane Structure

Which of the following correctly describes how the structure of a phospholipid bilayer contributes to its function as a selective barrier?
  1. A. The hydrophilic fatty acid tails face inward, blocking polar molecules from passing through
  2. B. Cholesterol molecules form a continuous impermeable layer between the two leaflets, preventing all molecule passage
  3. C. The phosphate heads embedded in the interior of the bilayer repel ions and maintain membrane integrity
  4. D. The hydrophobic core formed by nonpolar fatty acid tails prevents free passage of charged and polar molecules
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Correct answer: D

The two layers of nonpolar fatty acid tails face each other in the interior of the membrane, creating a hydrophobic core that repels charged ions and polar molecules, making the membrane selectively permeable. Choice A reverses the correct orientation — fatty acid tails are hydrophobic and face inward, while the phosphate heads are hydrophilic and face the aqueous environment.

Question 9 of 10 · Osmosis in Plant Cells

A plant cell is placed in a hypotonic solution. Which of the following correctly describes the expected outcome, and what structure prevents the cell from bursting?
  1. A. Water enters the cell; the cell wall provides structural rigidity that resists excessive expansion and prevents lysis
  2. B. Water leaves the cell; the plasma membrane prevents excessive water loss
  3. C. Water enters the cell; the central vacuole pumps excess water out to maintain volume
  4. D. Water enters the cell; the cell membrane bursts, just as in an animal cell
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Correct answer: A

In a hypotonic solution, water moves by osmosis into the plant cell (higher external water potential → lower internal water potential), increasing turgor pressure. The rigid cell wall resists expansion, preventing lysis. Animal cells lack a cell wall and do lyse under these conditions (making D incorrect for plant cells).

Question 10 of 10 · Membrane Transport and Osmosis

A researcher places red blood cells in a solution and observes that they swell and eventually lyse. Which of the following best explains this observation?
  1. A. The solution is hypertonic relative to the cell interior, causing water to exit by osmosis
  2. B. Solute molecules move from the solution into the cell, increasing internal pressure
  3. C. Active transport pumps concentrate solutes inside the cell, drawing water in against the gradient
  4. D. The solution is hypotonic relative to the cell interior, causing water to enter by osmosis
Show answer and explanation

Correct answer: D

When cells are placed in a hypotonic solution, the external solute concentration is lower than the cell interior, so water moves in by osmosis, causing swelling and lysis. A hypertonic solution (A) would cause the cell to shrink by losing water, not swell.

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