IB Biology Quiz: Apply Water Potential
20 questions · exam conditions
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Apply Water PotentialQuestion 1 of 20

A student observes a plasmolyzed plant cell under a microscope. Which statement correctly identifies the pressure potential (ψp) and describes the surrounding solution?

ψp is positive, and the solution is hypotonic.
ψp is zero, and the solution is hypotonic.
ψp is negative, and the solution is isotonic.
ψp is zero, and the solution is hypertonic.
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IB Biology Quiz

IB Biology Quiz: Apply Water Potential

Practice Apply Water Potential in IB Biology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Apply Water Potential, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Biology.

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Question 1

A student observes a plasmolyzed plant cell under a microscope. Which statement correctly identifies the pressure potential (ψp) and describes the surrounding solution?

  1. ψp is positive, and the solution is hypotonic.
  2. ψp is zero, and the solution is hypotonic.
  3. ψp is negative, and the solution is isotonic.
  4. ψp is zero, and the solution is hypertonic. (correct answer)
Explanation: Plasmolysis occurs when a plant cell loses a significant amount of water, causing the plasma membrane to pull away from the cell wall. This happens when the cell is placed in a hypertonic solution (a solution with a lower water potential). In this state, the cell is flaccid and exerts no outward pressure on the cell wall, so its pressure potential (turgor pressure) is zero.

Question 2

A turgid plant cell has a solute potential (ψs) of -1.2 MPa and is in equilibrium with its surroundings. The cell is then moved to a beaker of pure water. What will be the pressure potential (ψp) of the cell once it reaches a new equilibrium in the pure water?

  1. 0 MPa
  2. -1.2 MPa
  3. +1.2 MPa (correct answer)
  4. +2.4 MPa
Explanation: At equilibrium in pure water, the water potential (ψw) of the cell will equal the water potential of the pure water, which is 0 MPa. The water potential of the cell is the sum of its solute potential (ψs) and pressure potential (ψp), so ψw = ψs + ψp. Substituting the known values: 0 MPa = -1.2 MPa + ψp. Solving for ψp gives ψp = +1.2 MPa. This represents the maximum turgor pressure the cell can achieve.

Question 3

The water potential of the cytoplasm of a root hair cell is -0.6 MPa. The cell is absorbing water from the soil. Which of the following conditions in the soil would allow for the most rapid net movement of water into the root cell?

  1. Soil water potential of -0.6 MPa, due to isotonic conditions facilitating movement.
  2. Soil solute potential of -0.2 MPa and soil pressure potential of +0.1 MPa. (correct answer)
  3. Soil solute potential of -0.8 MPa and soil pressure potential of +0.1 MPa.
  4. Soil solute potential of -0.1 MPa and soil pressure potential of -0.1 MPa.
Explanation: Water moves down a water potential gradient, from a higher (less negative) water potential to a lower (more negative) water potential. The cell's water potential is -0.6 MPa. We need to find the soil condition with the highest water potential that is still greater than -0.6 MPa to allow for absorption. Choice A has no gradient. Choice B: soil ψw = -0.2 + 0.1 = -0.1 MPa. The gradient is (-0.1) - (-0.6) = 0.5 MPa. Choice C: soil ψw = -0.8 + 0.1 = -0.7 MPa; water would move out. Choice D: soil ψw = -0.1 + (-0.1) = -0.2 MPa. The gradient is (-0.2) - (-0.6) = 0.4 MPa. The gradient in B is the steepest, allowing for the most rapid net movement of water into the cell.

Question 4

A potato cylinder is placed in a sucrose solution and after one hour, its mass has not changed. Which statement correctly describes the potato cells at this point?

  1. The water potential of the potato cells is 0 MPa, and they are fully turgid.
  2. The water potential of the potato cells is equal to the water potential of the sucrose solution. (correct answer)
  3. The solute potential of the potato cells is equal to the solute potential of the sucrose solution.
  4. The net movement of water has stopped, and the cell membranes have become impermeable.
Explanation: If there is no change in mass, there is no net movement of water via osmosis. This state of equilibrium is reached when the water potential of the potato cells is equal to the water potential of the surrounding solution. Choice A is only true if the solution is pure water. Choice C is incorrect because the potato cells have pressure potential, so for ψw to be equal, ψs cannot be equal unless ψp is zero. Choice D is incorrect; water molecules continue to move across the membrane, but the rates in and out are equal, resulting in no net movement.

Question 5

The solute potential (ψs) can be calculated using the formula ψs = -iCRT, where 'i' is the ionization constant. A 0.3 M solution of glucose (which does not ionize) is found to be isotonic to a solution of calcium chloride (CaCl₂). What is the approximate molar concentration of the CaCl₂ solution?

  1. 0.1 M (correct answer)
  2. 0.2 M
  3. 0.3 M
  4. 0.9 M
Explanation: For the solutions to be isotonic, they must have the same solute potential, and therefore the same total concentration of solute particles. For glucose, i = 1, so the effective solute concentration is 1 × 0.3 M = 0.3 M. Calcium chloride (CaCl₂) dissociates into one Ca²⁺ ion and two Cl⁻ ions, so its ionization constant, i, is 3. To find the molar concentration (C) of CaCl₂ that gives an effective solute concentration of 0.3 M, we solve the equation: 3 × C = 0.3 M. Therefore, C = 0.1 M.

Question 6

Water moves from the cortex of a plant root into the xylem. For this to occur via osmosis, what must be true about the water potentials?

  1. ψw of cortex > ψw of xylem (correct answer)
  2. ψw of cortex < ψw of xylem
  3. ψs of cortex > ψs of xylem
  4. ψp of cortex < ψp of xylem
Explanation: The fundamental principle of water movement via osmosis is that water moves down a water potential gradient, from an area of higher water potential to an area of lower water potential. Therefore, for water to move from the cortex cells into the xylem, the water potential (ψw) of the cortex must be higher (less negative) than the water potential (ψw) of the xylem.

Question 7

A student investigating osmosis places dialysis tubing filled with a 0.5 M sucrose solution into a beaker of 0.2 M sucrose solution. The dialysis tubing is permeable to water but not to sucrose. What will be the net direction of water movement and the change in the tubing?

  1. Water will move out of the tubing, causing it to become flaccid.
  2. Water will move into the tubing, causing it to become more turgid. (correct answer)
  3. There will be no net movement of water as both solutions contain sucrose.
  4. Sucrose will move out of the tubing, causing the concentrations to equalize.
Explanation: The 0.5 M sucrose solution inside the tubing has a higher solute concentration and thus a lower (more negative) solute and water potential than the 0.2 M solution in the beaker. Water will move down its water potential gradient from the beaker into the dialysis tubing. This influx of water will increase the volume and internal pressure of the tubing, causing it to become more turgid.

Question 8

If the pressure potential (ψp) of a plant cell is +0.7 MPa and its water potential (ψw) is -0.2 MPa, what is the solute potential (ψs) of the cell?

  1. -0.9 MPa (correct answer)
  2. +0.5 MPa
  3. -0.5 MPa
  4. +0.9 MPa
Explanation: The water potential equation is ψw = ψs + ψp. To find the solute potential (ψs), we can rearrange the formula to ψs = ψw - ψp. Substituting the given values: ψs = (-0.2 MPa) - (+0.7 MPa) = -0.9 MPa. Solute potential is always negative or zero, which is consistent with the answer.

Question 9

Which of the following correctly pairs a biological fluid with its typical water potential relative to pure water?

  1. Human blood plasma: positive water potential.
  2. Seawater: water potential near zero.
  3. Xylem sap under tension: highly negative water potential. (correct answer)
  4. Pure water in a cell: zero water potential.
Explanation: Pure water has a water potential of 0 MPa. All biological fluids contain solutes, so their solute potentials are negative, making their overall water potentials negative. Xylem sap is under tension (negative pressure), which makes its pressure potential negative. This, combined with its negative solute potential, results in a highly negative water potential, which drives water transport. Blood plasma and seawater have negative water potentials due to solutes. Pure water can only exist in a cell if the cell is dead and has been washed out; a living cell has solutes and thus a negative water potential.

Question 10

Incipient plasmolysis is the point at which the plasma membrane just begins to pull away from the cell wall. A biologist determines that incipient plasmolysis occurs when a plant tissue is placed in a 0.4 M sucrose solution. What can be deduced about the plant cells?

  1. The pressure potential of the cells is at its maximum positive value.
  2. The water potential of the cells is equal to the solute potential of the 0.4 M sucrose solution. (correct answer)
  3. The solute potential of the cells is equal to the pressure potential of the cells.
  4. The water potential of the cells is 0 MPa.
Explanation: At incipient plasmolysis, the cell is flaccid, meaning it exerts no pressure on the cell wall. Therefore, its pressure potential (ψp) is zero. In this state, the cell's water potential is equal to its solute potential (ψw = ψs). Since the cell is at equilibrium with the surrounding solution (no net water movement at this specific point), its water potential must also be equal to the solution's water potential. The water potential of the sucrose solution is determined solely by its solute potential, as it is at atmospheric pressure. Therefore, the water potential (and solute potential) of the cells is equal to the solute potential of the 0.4 M sucrose solution.

Question 11

A freshwater protist, such as Paramecium, lives in a hypotonic environment. It uses a contractile vacuole to expel excess water. What would happen to the rate of vacuole contraction if the Paramecium was placed in a marine environment?

  1. The rate would increase because the external water potential is lower, driving more water in.
  2. The rate would decrease because the external environment is hypertonic or isotonic, reducing water influx. (correct answer)
  3. The rate would remain the same because the vacuole's function is independent of external water potential.
  4. The rate would increase to expel salts that diffuse into the cell from the marine water.
Explanation: In its native freshwater (hypotonic) environment, the Paramecium has a lower water potential than its surroundings, causing water to constantly enter by osmosis. The contractile vacuole works to expel this water. A marine environment is hypertonic, meaning its water potential is much lower than that of the protist's cytoplasm. This would reverse or greatly reduce the water potential gradient, causing water to move out of the cell, or enter at a much slower rate. Consequently, the need to expel excess water would diminish, and the rate of contractile vacuole contraction would decrease significantly.

Question 12

A plant cell with a water potential (ψw) of -0.9 MPa is placed in a solution with a water potential of -0.4 MPa. Which statement describes the initial state and subsequent net movement of water?

  1. The cell is hypertonic to the solution, and water will move into the cell. (correct answer)
  2. The cell is hypotonic to the solution, and water will move out of the cell.
  3. The cell is hypertonic to the solution, and water will move out of the cell.
  4. The cell is isotonic with the solution, and there will be no net movement of water.
Explanation: A lower (more negative) water potential indicates a higher effective solute concentration. Therefore, the cell (ψw = -0.9 MPa) is hypertonic to the solution (ψw = -0.4 MPa). Water always moves down a water potential gradient, from a region of higher water potential to a region of lower water potential. Since -0.4 MPa is higher than -0.9 MPa, water will move from the solution into the cell.

Question 13

Transpiration creates tension in the xylem, which is essential for pulling water up the stem. How is this tension correctly described in terms of water potential?

  1. A high positive solute potential (ψs) in the xylem sap.
  2. A large negative pressure potential (ψp) in the xylem. (correct answer)
  3. A pressure potential (ψp) of zero throughout the xylem column.
  4. A solute potential (ψs) that is less negative than the surrounding cells.
Explanation: Tension in a fluid column is a form of negative pressure. In the context of water potential, this is represented by a large negative pressure potential (ψp). This negative pressure potential makes the overall water potential (ψw) in the xylem extremely low (very negative), creating the gradient that pulls water up from the roots.

Question 14

An unpeeled grape is placed in a very salty solution. After several hours, the grape appears shrivelled. Which explanation is the most accurate?

  1. The high concentration of salt ions actively transported water out of the grape cells.
  2. The solute potential of the grape cells became less negative, causing them to lose water to the more negative solution.
  3. The pressure potential inside the grape cells increased, forcing water out into the lower-pressure solution.
  4. The water potential of the salt solution was lower than the water potential of the grape cells, causing a net loss of water. (correct answer)
Explanation: The salty solution is hypertonic to the grape cells. This means the solution has a high solute concentration and therefore a very low (highly negative) water potential. The grape cells have a higher water potential in comparison. Water moves passively by osmosis down the water potential gradient, from the grape cells into the surrounding solution, causing the cells to lose water and the grape to shrivel. Choice A is incorrect as water is not actively transported. Choice C describes the opposite of what happens to pressure potential. Choice D describes a change that would reduce water loss, not cause it.

Question 15

A flaccid plant cell with a solute potential of -1.0 MPa is placed in a solution with a water potential of -1.5 MPa. What will happen to the cell's water potential components?

  1. The cell will lose water, and its pressure potential will become positive.
  2. The cell will gain water, and its solute potential will become less negative.
  3. The cell will lose water, and its solute potential will become more negative. (correct answer)
  4. The cell will gain water, and its pressure potential will become positive.
Explanation: A flaccid cell has a pressure potential of zero, so its initial water potential is equal to its solute potential: ψw = -1.0 MPa. The solution's water potential is -1.5 MPa. Since -1.0 MPa > -1.5 MPa, water will move down the gradient from the cell to the solution. As the cell loses water, its cytoplasm becomes more concentrated, causing its solute potential (ψs) to become more negative. The cell will become plasmolyzed, and its pressure potential will remain at zero.

Question 16

Which of the following would cause the most significant decrease in the water potential of a solution in a beaker at atmospheric pressure?

  1. Increasing the temperature of the solution from 20 °C to 40 °C.
  2. Decreasing the atmospheric pressure above the beaker.
  3. Adding 10 grams of starch to the solution.
  4. Adding 10 grams of NaCl to the solution. (correct answer)
Explanation: Water potential (ψw) in a beaker is essentially equal to its solute potential (ψs), as pressure is atmospheric (zero). Solute potential becomes more negative as solute concentration increases. Both NaCl and starch are solutes, but for an equal mass, NaCl will have a much greater effect. This is because NaCl has a low molar mass (58.44 g/mol) and dissociates into two ions (Na⁺ and Cl⁻), greatly increasing the molar concentration of solute particles. Starch is a polymer with a very high molar mass, so 10 grams represents a very small number of moles, and it does not ionize. Temperature and atmospheric pressure have minor effects compared to the addition of a low-molar-mass, ionizable solute.

Question 17

An animal cell with an internal solute potential of -0.8 MPa is placed in a solution with a solute potential of -0.3 MPa. What is the predicted outcome?

  1. The cell will shrink as water moves out down the water potential gradient.
  2. The cell will swell and lyse as water moves in down the water potential gradient. (correct answer)
  3. The cell will remain unchanged as the pressure potential will balance the solute potential.
  4. The cell will initially swell but then return to its original size via active transport of water.
Explanation: Animal cells have no cell wall, so their pressure potential is assumed to be zero. The water potential of the cell is equal to its solute potential, ψw(cell) = -0.8 MPa. The water potential of the solution is ψw(solution) = -0.3 MPa. Since -0.3 MPa is higher (less negative) than -0.8 MPa, water will move down the gradient from the solution into the cell. Without a cell wall to build turgor pressure, the cell will continue to swell until it bursts (lyses).

Question 18

During the opening of stomata, guard cells actively transport potassium ions (K⁺) into their cytoplasm. What is the immediate consequence of this action in terms of water potential?

  1. The pressure potential of the guard cells decreases, causing water to exit.
  2. The solute potential of the guard cells becomes more negative, causing water to enter. (correct answer)
  3. The water potential of the guard cells becomes less negative, causing water to enter.
  4. The solute potential of the guard cells becomes less negative, causing water to exit.
Explanation: The active transport of K⁺ ions into the guard cells increases the solute concentration within them. An increase in solute concentration makes the solute potential (ψs) more negative. This lowers the overall water potential (ψw) of the guard cells, creating a gradient that causes water to move from the surrounding epidermal cells into the guard cells via osmosis.

Question 19

In the phloem, sucrose is actively loaded into sieve-tube elements at the source. How does this facilitate the mass flow of sap?

  1. It increases the pressure potential in the sieve-tube, forcing water into the adjacent xylem.
  2. It makes the solute potential in the sieve-tube more negative, drawing water in from the xylem and increasing pressure. (correct answer)
  3. It makes the water potential of the xylem more negative, pushing water into the phloem.
  4. It requires ATP, which directly pumps the phloem sap from the source to the sink.
Explanation: The active loading of sucrose into the sieve-tube elements increases their internal solute concentration. This makes their solute potential (ψs) more negative, which in turn lowers their overall water potential (ψw). This creates a water potential gradient between the phloem and the nearby xylem. Water moves by osmosis from the xylem (higher ψw) into the sieve-tube elements (lower ψw), generating a high positive pressure potential (turgor pressure). This high pressure drives the bulk flow of sap towards the sink.

Question 20

Why can a plant cell become turgid while an animal cell lyses when both are placed in pure water?

  1. Plant cells have a higher initial solute potential than animal cells, preventing excessive water entry.
  2. Plant cells can actively pump out excess water using their large central vacuole, a mechanism absent in animal cells.
  3. The animal cell membrane is more permeable to water, allowing for a faster influx that the cell cannot handle.
  4. The plant cell wall exerts an opposing pressure potential that increases as water enters, eventually stopping net influx. (correct answer)
Explanation: When a plant cell is placed in pure water, water enters due to the low internal water potential. As water enters, the cell swells and the plasma membrane pushes against the rigid cell wall. The cell wall pushes back, creating a positive pressure potential (turgor pressure). This pressure potential raises the cell's overall water potential. Equilibrium is reached when the cell's water potential rises to 0 MPa (equal to pure water), at which point net water movement ceases. An animal cell lacks a cell wall and cannot build significant pressure potential; water influx continues until the membrane ruptures (lysis).