AP Biology Quiz: Tonicity And Osmoregulation
20 questions · exam conditions
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Tonicity And OsmoregulationQuestion 1 of 20

Two groups of identical animal cells (internal solute concentration 0.25 M) are placed into separate solutions. Group X is placed in 0.25 M glucose; Group Y is placed in 0.25 M sucrose. The membrane is permeable to water and glucose but not to sucrose. After 15 minutes, Group X cells are larger than at the start, while Group Y cells show little change. Which outcome is most likely to explain why Group X swelled?

Glucose enters the cells, increasing internal solute concentration and drawing water inward.
Sucrose enters the cells, increasing internal solute concentration and drawing water inward.
Water leaves Group X because glucose lowers external water concentration more than sucrose.
Water enters Group Y because sucrose diffuses through the membrane faster than glucose.
No solute movement occurs; Group X swells because isotonic solutions cause net water influx.
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AP Biology Quiz

AP Biology Quiz: Tonicity And Osmoregulation

Practice Tonicity And Osmoregulation in AP Biology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Tonicity And Osmoregulation, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Biology.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Two groups of identical animal cells (internal solute concentration 0.25 M) are placed into separate solutions. Group X is placed in 0.25 M glucose; Group Y is placed in 0.25 M sucrose. The membrane is permeable to water and glucose but not to sucrose. After 15 minutes, Group X cells are larger than at the start, while Group Y cells show little change. Which outcome is most likely to explain why Group X swelled?

  1. Glucose enters the cells, increasing internal solute concentration and drawing water inward. (correct answer)
  2. Sucrose enters the cells, increasing internal solute concentration and drawing water inward.
  3. Water leaves Group X because glucose lowers external water concentration more than sucrose.
  4. Water enters Group Y because sucrose diffuses through the membrane faster than glucose.
  5. No solute movement occurs; Group X swells because isotonic solutions cause net water influx.

Explanation: This question assesses understanding of tonicity and osmoregulation when membranes are permeable to some solutes. In Group X, glucose permeates the membrane, entering cells and raising internal solute concentration above 0.25 M, lowering internal water potential. Water then moves from higher external potential to lower internal, causing swelling. Group Y remains isotonic since sucrose cannot enter, highlighting permeability's role. A tempting distractor is choice B, suggesting sucrose enters similarly, but this ignores membrane selectivity, a misconception overlooking differential permeability. A transferable strategy is to factor in solute permeability when calculating effective tonicity to predict dynamic changes over time.

Question 2

A student places identical animal cells (internal solute concentration 0.30 M) into three beakers separated by membranes permeable only to water. Beaker 1 contains 0.10 M sucrose, Beaker 2 contains 0.30 M sucrose, and Beaker 3 contains 0.50 M sucrose. After 10 minutes, the student observes no change in the cells in Beaker 2. Assume sucrose does not cross the cell membrane and no active transport changes solute levels during the trial. Which outcome is most likely for the cells in Beaker 3 compared with their initial volume?

  1. They swell because water moves into the cells down the solute gradient.
  2. They shrink because water moves out of the cells toward the higher solute concentration. (correct answer)
  3. They remain unchanged because equal solute concentrations prevent net water movement.
  4. They lyse because sucrose diffuses into the cells, increasing internal solute concentration.
  5. They shrink because cells actively pump water out to maintain a constant volume.

Explanation: This question assesses understanding of tonicity and osmoregulation, which involve how cells maintain water balance in different solute environments. The cells in Beaker 3 are exposed to a 0.50 M sucrose solution, which has a higher solute concentration than the internal 0.30 M, resulting in lower water potential outside the cells. Water potential drives water movement from areas of higher potential (inside the cells) to lower potential (outside), causing net water efflux and cell shrinkage. This hypertonic condition explains why the cells shrink compared to their initial volume, as observed similarly with no change in the isotonic Beaker 2. A tempting distractor is choice A, which suggests swelling due to water influx, but this confuses hypertonic with hypotonic environments, a common misconception where tonicity directions are reversed. To approach similar problems, always compare solute concentrations to determine tonicity and predict water movement based on water potential gradients.

Question 3

A cell's cytosol contains 0.28 M nonpenetrating solute. It is moved to a solution with 0.12 M nonpenetrating solute, and the membrane is permeable only to water. Soon after transfer, the cell's volume increases. Which outcome is most likely?

  1. Net water moved out because the external solution was hypotonic to the cytosol.
  2. Net water moved in because the external solution was hypotonic to the cytosol. (correct answer)
  3. No net water movement occurred because the solute could not cross the membrane.
  4. Net water moved out because the external solution had lower solute concentration.
  5. Net water moved in because solute was actively transported into the cell.

Explanation: This question tests tonicity and osmoregulation by analyzing water movement in response to solute gradients. The cell's cytosol contains 0.28 M solute and is placed in a 0.12 M external solution, making the external solution hypotonic (lower solute concentration) relative to the cytosol. Water moves by osmosis from the area of higher water potential (outside, with 0.12 M solute) to the area of lower water potential (inside, with 0.28 M solute). This water influx causes the cell volume to increase, which matches the observed swelling. Choice A incorrectly states that water moved out when the external solution was hypotonic, which would actually cause water to move in. Always remember that hypotonic solutions have lower solute concentration and cause cells to gain water and swell.

Question 4

A cell with 0.20 M nonpenetrating solute inside is placed in a solution containing 0.05 M nonpenetrating solute. The membrane is permeable to water but not to the solute. After several minutes, the cell's volume increases. Which outcome is most likely?

  1. Water moved into the cell because the external solution was hypotonic relative to the cytosol. (correct answer)
  2. Water moved out of the cell because the external solution was hypotonic relative to the cytosol.
  3. Water moved out of the cell because the external solution was isotonic relative to the cytosol.
  4. No net water movement occurred because solute concentration does not affect osmosis.
  5. Water moved into the cell because solute diffused into the cell, increasing internal solute.

Explanation: This question tests tonicity and osmoregulation concepts by examining water movement across a selectively permeable membrane. The cell contains 0.20 M solute internally and is placed in a 0.05 M external solution, making the external solution hypotonic (lower solute concentration) relative to the cytosol. Water moves osmotically from the area of higher water potential (outside, with 0.05 M solute) to the area of lower water potential (inside, with 0.20 M solute). This water influx causes the cell volume to increase, confirming the hypotonic condition. Choice D incorrectly claims that solute concentration doesn't affect osmosis, when in fact solute concentration differences drive osmotic water movement. Always compare solute concentrations to predict water movement: water flows from hypotonic (dilute) to hypertonic (concentrated) solutions.

Question 5

A student fills dialysis tubing (permeable to water but not to starch) with 0.60 M starch solution and places it into a beaker containing 0.20 M starch solution. The initial mass of the tubing is recorded, and after 20 minutes the tubing has increased in mass. Assume temperature and pressure remain constant and no starch crosses the tubing. Which outcome is most likely responsible for the mass change?​

  1. Net water movement into the tubing because the beaker solution is hypertonic to the tubing.
  2. Net water movement out of the tubing because the tubing solution has higher water potential.
  3. Net water movement into the tubing because the tubing solution is hypertonic to the beaker. (correct answer)
  4. Net starch movement into the tubing because starch diffuses down its concentration gradient.
  5. No net water movement because both solutions contain starch, so they are isotonic.

Explanation: This question assesses understanding of tonicity and osmoregulation using dialysis tubing as a model for semipermeable membranes. The tubing's 0.60 M starch solution has higher solute concentration than the beaker's 0.20 M, making the inside hypertonic with lower water potential internally. Water moves from higher potential in the beaker to lower inside the tubing, increasing mass due to net influx. This osmosis occurs without starch diffusion, explaining the observed change. A tempting distractor is choice A, claiming influx because the beaker is hypertonic, but this reverses the tonicity comparison, a misconception arising from confusing which side has higher solutes. For a transferable strategy, visualize water potential as 'pulling' water toward higher solute areas to anticipate mass or volume changes in enclosed systems.

Question 6

Plant cells with an internal solute concentration of 0.40 M are placed into a solution of 0.20 M solute. The plasma membrane is permeable to water but not to the solute, and the cell wall is intact. After several minutes, the central vacuole appears larger and the plasma membrane presses against the cell wall. Assume no solute transport occurs during this time. Which direction of net water movement best explains these observations?

  1. Water moves into the cells because the external solution is hypotonic to the cytoplasm. (correct answer)
  2. Water moves out of the cells because the external solution is hypertonic to the cytoplasm.
  3. Water moves into the cells because solute diffuses into the vacuole, increasing water potential.
  4. No net water movement occurs because the cell wall prevents osmosis across the membrane.
  5. No net water movement occurs because the external solute concentration is lower than internal.

Explanation: This question assesses understanding of tonicity and osmoregulation, focusing on how plant cells respond to external solute concentrations. The external 0.20 M solution has lower solute concentration than the internal 0.40 M, creating a hypotonic environment with higher water potential outside. Water moves from higher potential outside to lower potential inside, increasing the central vacuole size and causing turgor pressure against the cell wall. This net influx explains the observed expansion without solute crossing the membrane. A tempting distractor is choice B, which claims water moves out in a hypertonic solution, but this misinterprets the tonicity by swapping hypotonic and hypertonic conditions, a frequent misconception. For transferable strategy, calculate water potential differences by comparing solute molarities to predict osmosis direction in cells with rigid walls.

Question 7

A student observes plant cells under a microscope. In Solution A (0.55 M solute), the plasma membrane pulls away from the cell wall. In Solution B (0.15 M solute), the vacuole expands and the membrane presses against the wall. The membrane is permeable to water but not the solute. Which statement best predicts net water movement in Solution A?

  1. Water moves into the cells because Solution A has a higher water concentration than the cytoplasm.
  2. Water moves out of the cells because Solution A is hypertonic relative to the cytoplasm. (correct answer)
  3. Water moves out of the cells because Solution A is hypotonic relative to the cytoplasm.
  4. No net water movement occurs because plasmolysis blocks osmosis across the membrane.
  5. No net water movement occurs because the cell wall prevents changes in cell water content.

Explanation: This question assesses understanding of tonicity and osmoregulation through observed plant cell responses like plasmolysis. Solution A at 0.55 M has higher solute concentration than the cytoplasm, creating a hypertonic environment with lower external water potential. Water moves out from higher internal potential to lower external, causing the plasma membrane to pull away (plasmolysis). This contrasts with turgor in the hypotonic Solution B, confirming tonicity effects. A tempting distractor is choice C, predicting efflux but labeling it hypotonic, which confuses tonicity terms, a common misconception in interpreting observations. For a transferable strategy, use visual cues like plasmolysis or turgor to infer tonicity and apply water potential logic to unseen scenarios.

Question 8

A student places a cell with cytosol at 0.30 M nonpenetrating solute into a solution at 0.60 M nonpenetrating solute. The membrane is permeable to water but not solute, and the external solution stays at 0.60 M. The cell starts at normal volume and is observed for 2 minutes. Which outcome is most likely regarding net water movement?

  1. Net water movement is into the cell because water moves toward lower solute concentration.
  2. Net water movement is out of the cell because the external solution is hypertonic. (correct answer)
  3. Net water movement is zero because solute cannot cross the membrane.
  4. Net water movement is into the cell because the external solution has higher solute concentration.
  5. Net water movement is out of the cell only if aquaporins use ATP to transport water.

Explanation: This question assesses the skill of tonicity and osmoregulation, determining net water direction in hypertonic scenarios. The 0.60 M solution is hypertonic to the 0.30 M cell, creating lower external water potential that pulls water out. Net movement is thus out of the cell as osmosis seeks to balance concentrations. This would lead to cell dehydration if prolonged. Choice D is tempting, stating influx because of higher external solute, but this confuses the rule, from the misconception that water moves toward higher solute rather than lower water potential. Focus on water potential gradients to predict net movement in osmosis problems effectively.

Question 9

Two beakers contain nonpenetrating solute solutions: Beaker 1 is 0.20 M and Beaker 2 is 0.50 M. Identical cells have an internal solute concentration of 0.35 M and membranes permeable to water but not solute. Cells are placed into each beaker and observed for 5 minutes; external concentrations stay constant. Which outcome is most likely for cell volume in Beaker 1 compared with Beaker 2?

  1. Cells in Beaker 1 shrink more than cells in Beaker 2 because 0.20 M is hypertonic.
  2. Cells in Beaker 1 swell, while cells in Beaker 2 shrink, due to opposite tonicity. (correct answer)
  3. Cells in both beakers swell because water always moves into cells in solute solutions.
  4. Cells in both beakers remain unchanged because the solute cannot cross the membrane.
  5. Cells in Beaker 1 shrink, while cells in Beaker 2 swell, due to water moving down its gradient.

Explanation: This question assesses the skill of tonicity and osmoregulation, comparing cell behaviors in varying tonicities. Beaker 1 at 0.20 M is hypotonic to the 0.35 M cells, so higher external water potential drives water in, causing swelling. Beaker 2 at 0.50 M is hypertonic, with lower external water potential pulling water out, leading to shrinking. These opposite effects arise from water moving down its potential gradient in each case. Choice A tempts by suggesting Beaker 1 shrinks more, but this reverses tonicity, due to the misconception that lower external solute is hypertonic rather than hypotonic. To solve comparatives, label each solution's tonicity relative to the cell and map water flow accordingly.

Question 10

Animal cells (0.35 M internal nonpenetrating solute) are placed in a 0.35 M nonpenetrating solute solution. Water can cross the membrane, but solute cannot. After 10 minutes, the cells look unchanged. Which outcome is most likely?

  1. Net water moved into the cells because the outside solution was hypotonic.
  2. Net water moved out of the cells because the outside solution was hypertonic.
  3. No net water movement occurred because the solutions were isotonic. (correct answer)
  4. Net water moved into the cells because solute moved out, lowering cytosolic solute.
  5. No net water movement occurred because water cannot move without ATP.

Explanation: This question evaluates tonicity and osmoregulation when cells are in equilibrium with their environment. The animal cells contain 0.35 M internal solute and are placed in a 0.35 M external solution, creating isotonic conditions where solute concentrations are equal inside and outside. Under isotonic conditions, water potential is the same on both sides of the membrane, resulting in no net water movement—water molecules move equally in both directions. This equilibrium explains why the cells appear unchanged after 10 minutes. Choice E incorrectly suggests that water movement requires ATP, when osmosis is actually a passive process driven by concentration gradients. To identify isotonic conditions, compare internal and external solute concentrations; when equal, expect no net water movement or cell size change.

Question 11

A student fills dialysis tubing (permeable to water but not to starch) with 0.60 M starch solution and places it into a beaker containing 0.20 M starch solution. The initial mass of the tubing is recorded, and after 20 minutes the tubing has increased in mass. Assume temperature and pressure remain constant and no starch crosses the tubing. Which outcome is most likely responsible for the mass change?

  1. Net water movement into the tubing because the beaker solution is hypertonic to the tubing.
  2. Net water movement out of the tubing because the tubing solution has higher water potential.
  3. Net water movement into the tubing because the tubing solution is hypertonic to the beaker. (correct answer)
  4. Net starch movement into the tubing because starch diffuses down its concentration gradient.
  5. No net water movement because both solutions contain starch, so they are isotonic.

Explanation: This question assesses understanding of tonicity and osmoregulation using dialysis tubing as a model for semipermeable membranes. The tubing's 0.60 M starch solution has higher solute concentration than the beaker's 0.20 M, making the inside hypertonic with lower water potential internally. Water moves from higher potential in the beaker to lower inside the tubing, increasing mass due to net influx. This osmosis occurs without starch diffusion, explaining the observed change. A tempting distractor is choice A, claiming influx because the beaker is hypertonic, but this reverses the tonicity comparison, a misconception arising from confusing which side has higher solutes. For a transferable strategy, visualize water potential as 'pulling' water toward higher solute areas to anticipate mass or volume changes in enclosed systems.

Question 12

A cell with 0.10 M nonpenetrating solute inside is placed into a 0.20 M nonpenetrating solute solution. The membrane is permeable to water but not to the solute. After 8 minutes, the cell's volume decreases. Which outcome is most likely?

  1. Water moved into the cell because the external solution was hypotonic to the cytosol.
  2. Water moved out of the cell because the external solution was hypertonic to the cytosol. (correct answer)
  3. No net water movement occurred because the solute cannot cross the membrane.
  4. Water moved into the cell because the external solution had higher solute concentration.
  5. Water moved out of the cell because the cell actively transported water outward.

Explanation: This question assesses tonicity and osmoregulation by examining water movement when cells are placed in a more concentrated solution. The cell contains 0.10 M solute internally and is placed in a 0.20 M external solution, making the external solution hypertonic (higher solute concentration) relative to the cytosol. Water moves osmotically from the area of higher water potential (inside the cell, with lower solute) to the area of lower water potential (outside the cell, with higher solute). This water loss causes the cell volume to decrease, which matches the observed shrinkage. Choice D incorrectly states that water moves into the cell because of higher external solute, when actually water moves out toward the higher solute concentration. Remember that water always moves toward the hypertonic solution, causing cells in hypertonic environments to lose water and shrink.

Question 13

A cell with 0.50 M internal nonpenetrating solute is placed in 0.75 M nonpenetrating solute. Water can cross the membrane, but solute cannot. After several minutes, the cell's volume decreases. Which outcome is most likely for the direction of net water movement?

  1. Net water moved into the cell because the external solution was hypotonic.
  2. Net water moved out of the cell because the external solution was hypertonic. (correct answer)
  3. No net water movement occurred because both solutions contained solute.
  4. Net water moved into the cell because water moves toward higher solute inside.
  5. Net water moved out of the cell because solute diffused into the cytosol.

Explanation: This question examines tonicity and osmoregulation when cells face a hypertonic environment. The cell contains 0.50 M internal solute and is placed in a 0.75 M external solution, making the external solution hypertonic (higher solute concentration) relative to the cytosol. Water moves osmotically from the area of higher water potential (inside the cell, with lower solute concentration) to the area of lower water potential (outside the cell, with higher solute concentration). This water efflux causes the cell volume to decrease, which matches the observed shrinkage. Choice D incorrectly suggests water moves into the cell toward higher internal solute, when water actually moves out toward the even higher external solute concentration. To predict osmotic water movement, always identify which solution has higher solute concentration—water moves toward that hypertonic solution.

Question 14

A student places identical cells (internal solute concentration 0.20 M) into a solution of 0.05 M solute. The membrane is permeable to water but not to the solute. After 8 minutes, the average cell volume increases. Which outcome is most likely if the same cells are instead placed into a 0.80 M solute solution?

  1. Average volume increases because the external solution has lower water potential than the cytoplasm.
  2. Average volume decreases because the external solution is hypertonic, causing net water efflux. (correct answer)
  3. Average volume stays constant because water moves equally in both directions in any solution.
  4. Average volume increases because solute moves into cells, increasing internal water concentration.
  5. Average volume decreases because cells actively pump water out against the gradient.

Explanation: This question assesses understanding of tonicity and osmoregulation by contrasting hypotonic and hypertonic outcomes. The 0.80 M solution surpasses the cells' 0.20 M, creating a hypertonic setup with lower external water potential. Water flows out from higher internal to lower external potential, decreasing volume. This opposes the observed increase in the hypotonic 0.05 M solution. A tempting distractor is choice A, claiming volume increase due to lower external water potential, but this confuses hypertonic influx with efflux, a misconception in potential gradients. A transferable strategy is to contrast with known hypotonic results to deduce hypertonic behaviors using consistent water potential rules.

Question 15

A student places animal cells with an internal solute concentration of 0.30 M into a beaker containing 0.10 M nonpenetrating solute. The plasma membrane is permeable to water but not to the solute. After 10 minutes, the cells appear swollen compared with their initial size. Which outcome is most likely for net water movement across the membrane during the 10 minutes?

  1. Water moved into the cells because the external solution was hypotonic to the cytosol. (correct answer)
  2. Water moved out of the cells because the external solution was hypertonic to the cytosol.
  3. Water moved into the cells because solute was actively transported into the cytosol.
  4. No net water movement occurred because the solute concentrations were equal.
  5. Water moved out of the cells because water concentration was lower outside the cells.

Explanation: This question tests your understanding of tonicity and osmoregulation, specifically how water moves across cell membranes in response to solute concentration differences. The animal cells have an internal solute concentration of 0.30 M and are placed in a 0.10 M solution, making the external solution hypotonic (lower solute concentration) relative to the cytosol. Water moves by osmosis from areas of higher water potential (lower solute concentration) to areas of lower water potential (higher solute concentration), so water flows into the cells. This influx of water causes the cells to swell, which matches the observed outcome. Choice B incorrectly identifies the solution as hypertonic, which would cause shrinking, not swelling. To solve tonicity problems, always compare solute concentrations: if external < internal, the solution is hypotonic and water enters the cell.

Question 16

In an experiment, cells with internal solute concentration 0.45 M are placed into Solution A (0.30 M) and Solution B (0.60 M). The membrane is permeable to water but not to the solute. After 10 minutes, cells in Solution A have increased in volume. Which outcome is most likely for cells placed in Solution B over the same time interval?

  1. They increase in volume because Solution B has a lower solute concentration than the cytoplasm.
  2. They decrease in volume because Solution B is hypertonic relative to the cytoplasm. (correct answer)
  3. They remain unchanged because water cannot cross membranes without solute transport.
  4. They increase in volume because Solution B has higher solute concentration, attracting water inward.
  5. They decrease in volume because solute enters the cells, raising internal water concentration.

Explanation: This question assesses understanding of tonicity and osmoregulation by predicting based on prior observations. Solution B at 0.60 M exceeds the cells' 0.45 M, establishing a hypertonic condition with lower external water potential. Water departs from higher internal to lower external potential, reducing volume. This mirrors the inverse of expansion in hypotonic Solution A at 0.30 M. A tempting distractor is choice A, suggesting increase due to lower solute externally, but this inverts hypotonic and hypertonic, a common misconception in gradient interpretation. A transferable strategy is to leverage patterns from one tonicity to infer opposites, applying water potential universally across experiments.

Question 17

A cell has an internal solute concentration of 0.35 M. It is placed into Solution 1 (0.35 M), and its volume does not change. It is then placed into Solution 2 (0.50 M), and its volume decreases. The membrane is permeable to water but not to the solute. Which inference about Solution 2 is most consistent with the observation?

  1. Solution 2 is hypotonic to the cell, so water enters and the cell shrinks.
  2. Solution 2 is isotonic to the cell, so water leaves and the cell shrinks.
  3. Solution 2 is hypertonic to the cell, so water leaves and the cell shrinks. (correct answer)
  4. Solution 2 is hypertonic to the cell, so solute enters and the cell shrinks.
  5. Solution 2 is hypotonic to the cell, so solute leaves and the cell shrinks.

Explanation: This question assesses understanding of tonicity and osmoregulation by inferring solution properties from volume changes. Solution 2 at 0.50 M exceeds the cell's 0.35 M, making it hypertonic with lower water potential externally. Water exits from higher internal potential to lower external, resulting in shrinkage. This follows the no-change isotonic control in Solution 1. A tempting distractor is choice A, suggesting hypotonic conditions cause shrinkage, but this misapplies tonicity effects, a misconception from reversing water movement directions. As a transferable strategy, use control observations like isotonic stability to benchmark and predict outcomes in varied concentrations.

Question 18

A lab group places identical cells (cytosol 0.20 M nonpenetrating solute) into three solutions: I = 0.10 M, II = 0.20 M, III = 0.35 M. Membranes are permeable to water but not solute, and each solution's concentration stays constant. The group records whether cells swell, shrink, or stay the same after 5 minutes. Which result is most likely for cells in Solution III?

  1. Cells swell because 0.35 M has higher water concentration than the cytosol.
  2. Cells shrink because 0.35 M is hypertonic relative to the cytosol. (correct answer)
  3. Cells stay the same because Solution III is isotonic relative to the cytosol.
  4. Cells swell because solute enters the cell and increases internal osmolarity.
  5. Cells shrink only if the cells actively pump water out across the membrane.

Explanation: This question assesses the skill of tonicity and osmoregulation, predicting outcomes in a series of solutions. For Solution III at 0.35 M against 0.20 M cytosol, it's hypertonic with lower external water potential, drawing water out and causing shrinking. This efflux reduces cell volume as water moves to the lower potential. The nonpenetrating solute ensures the gradient persists without equalization. Choice A tempts by suggesting swelling from higher external water concentration, but this misreads the gradient, due to the misconception that higher solute externally means higher water there. Classify each solution's tonicity individually and use water potential to forecast changes consistently.

Question 19

Red blood cells (internal solute concentration 300 mOsm) are placed into a solution of 300 mOsm NaCl for 5 minutes, then transferred to a 150 mOsm NaCl solution. The membrane is permeable to water but not to NaCl. Immediately after transfer, the cells begin changing volume. Which outcome is most likely after several minutes in the 150 mOsm solution?

  1. Cells shrink because the outside has lower water concentration than the cytoplasm.
  2. Cells swell because the outside solution is hypotonic, causing net water influx. (correct answer)
  3. Cells remain unchanged because NaCl cannot cross the membrane, preventing osmosis.
  4. Cells remain unchanged because they were previously in an isotonic solution.
  5. Cells shrink because NaCl diffuses out, lowering external solute concentration.

Explanation: This question assesses understanding of tonicity and osmoregulation in red blood cells transferred between solutions. The 150 mOsm NaCl solution has lower solute concentration than the internal 300 mOsm, making it hypotonic with higher external water potential. Water flows from higher potential outside to lower inside, causing net influx and cell swelling. This shift from isotonic to hypotonic explains the volume increase after transfer. A tempting distractor is choice A, suggesting shrinkage due to lower external water concentration, but this inverts the relationship between solute and water concentrations, a common misconception in osmosis. As a transferable strategy, remember that water moves toward higher solute concentrations to dilute them, helping predict outcomes in changing environments.

Question 20

A student places identical animal cells (internal solute concentration 0.30 M) into Solution X (0.10 M nonpenetrating solute). After 5 minutes, which outcome is most likely for the cells? Assume membranes are permeable to water but not the solute, and temperature is constant. Water movement is driven by differences in solute concentration across the plasma membrane. The cells begin at the same volume and have no cell wall. Ignore any active transport and focus only on osmosis. The external solution remains at 0.10 M throughout the observation period.

  1. Cells gain water and swell because the surrounding solution is hypotonic to the cytosol. (correct answer)
  2. Cells lose water and shrink because the surrounding solution is hypertonic to the cytosol.
  3. Cells remain unchanged because equal solute concentrations prevent net water movement.
  4. Cells swell because solute diffuses into the cells and pulls water inward.
  5. Cells shrink because water is actively pumped out to maintain internal concentration.

Explanation: This question assesses the skill of tonicity and osmoregulation, focusing on how cells respond to different solute environments through osmosis. The cell's internal solute concentration is 0.30 M, while Solution X is 0.10 M, making it hypotonic with higher water potential outside than inside. Water moves from high to low water potential, so net water influx occurs, causing the cells to gain water and swell. This is because the lower external solute concentration creates a gradient driving water into the cytosol to dilute the higher internal concentration. A tempting distractor is choice D, which suggests cells swell due to solute diffusion inward, but this is wrong due to the misconception that nonpenetrating solutes can cross the membrane, whereas they cannot, and osmosis is driven by water potential, not solute movement. To approach similar problems, always compare solute concentrations to determine tonicity and predict water movement direction.