Tonicity and Osmoregulation
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AP Biology › Tonicity and Osmoregulation
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?
No solute movement occurs; Group X swells because isotonic solutions cause net water influx.
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.
Glucose enters the cells, increasing internal solute concentration and drawing water inward.
Water enters Group Y because sucrose diffuses through the membrane faster than glucose.
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.
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?
They remain unchanged because equal solute concentrations prevent net water movement.
They shrink because cells actively pump water out to maintain a constant volume.
They lyse because sucrose diffuses into the cells, increasing internal solute concentration.
They shrink because water moves out of the cells toward the higher solute concentration.
They swell because water moves into the cells down the solute gradient.
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.
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?
Net water moved in because solute was actively transported into the cell.
Net water moved in because the external solution was hypotonic to the cytosol.
Net water moved out because the external solution was hypotonic to the cytosol.
Net water moved out because the external solution had lower solute concentration.
No net water movement occurred because the solute could not cross the membrane.
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.
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?
Water moved out of the cell because the external solution was hypotonic relative to the cytosol.
Water moved out of the cell because the external solution was isotonic relative to the cytosol.
Water moved into the cell because the external solution was hypotonic relative to the cytosol.
No net water movement occurred because solute concentration does not affect osmosis.
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.
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?
Net water movement into the tubing because the tubing solution is hypertonic to the beaker.
Net water movement into the tubing because the beaker solution is hypertonic to the tubing.
Net water movement out of the tubing because the tubing solution has higher water potential.
No net water movement because both solutions contain starch, so they are isotonic.
Net starch movement into the tubing because starch diffuses down its concentration gradient.
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.
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?
Water moves into the cells because solute diffuses into the vacuole, increasing water potential.
Water moves out of the cells because the external solution is hypertonic to the cytoplasm.
No net water movement occurs because the cell wall prevents osmosis across the membrane.
No net water movement occurs because the external solute concentration is lower than internal.
Water moves into the cells because the external solution is hypotonic to the cytoplasm.
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.
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?
Water moves out of the cells because Solution A is hypotonic relative to the cytoplasm.
Water moves out of the cells because Solution A is hypertonic relative to the cytoplasm.
No net water movement occurs because plasmolysis blocks osmosis across the membrane.
Water moves into the cells because Solution A has a higher water concentration than the cytoplasm.
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.
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?
Net water movement is into the cell because water moves toward lower solute concentration.
Net water movement is zero because solute cannot cross the membrane.
Net water movement is out of the cell because the external solution is hypertonic.
Net water movement is into the cell because the external solution has higher solute concentration.
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.
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?
Cells in both beakers swell because water always moves into cells in solute solutions.
Cells in both beakers remain unchanged because the solute cannot cross the membrane.
Cells in Beaker 1 swell, while cells in Beaker 2 shrink, due to opposite tonicity.
Cells in Beaker 1 shrink, while cells in Beaker 2 swell, due to water moving down its gradient.
Cells in Beaker 1 shrink more than cells in Beaker 2 because 0.20 M is hypertonic.
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.
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?
Net water moved into the cells because the outside solution was hypotonic.
Net water moved into the cells because solute moved out, lowering cytosolic solute.
No net water movement occurred because water cannot move without ATP.
Net water moved out of the cells because the outside solution was hypertonic.
No net water movement occurred because the solutions were isotonic.
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.