IB Biology Quiz: Understand Water Potential
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Understand Water PotentialQuestion 1 of 20

A plasmolyzed plant cell with ψw = -1.5 MPa and ψp = 0 MPa is moved to pure water. Which sequence best describes the process of deplasmolysis?

Water enters the cell, causing ψp to increase from 0, making ψw less negative until it reaches equilibrium at 0 MPa.
Water enters the cell, causing ψs to become less negative due to dilution, which is the primary factor raising ψw to 0 MPa.
The cell wall actively pumps in water, increasing ψp and ψw until they are both positive.
Solutes exit the cell to balance the external solution, causing ψs to approach 0 MPa.
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IB Biology Quiz

IB Biology Quiz: Understand Water Potential

Practice Understand 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.

What this quiz covers

This quiz focuses on Understand 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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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.

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Question 1

A plasmolyzed plant cell with ψw = -1.5 MPa and ψp = 0 MPa is moved to pure water. Which sequence best describes the process of deplasmolysis?

  1. Water enters the cell, causing ψp to increase from 0, making ψw less negative until it reaches equilibrium at 0 MPa. (correct answer)
  2. Water enters the cell, causing ψs to become less negative due to dilution, which is the primary factor raising ψw to 0 MPa.
  3. The cell wall actively pumps in water, increasing ψp and ψw until they are both positive.
  4. Solutes exit the cell to balance the external solution, causing ψs to approach 0 MPa.
Explanation: When the plasmolyzed cell is placed in pure water (ψw = 0 MPa), there's a large water potential gradient. Water moves into the cell via osmosis. This increases the cell's volume, causing the plasma membrane to press against the cell wall. This generates turgor pressure, so the pressure potential (ψp) increases from 0 to a positive value. This increase in ψp directly raises the cell's overall water potential (ψw = ψs + ψp). The process continues until the cell's ψw equals the external ψw of 0 MPa.

Question 2

A plant cell has a solute potential (ψs) of -0.8 MPa and a pressure potential (ψp) of +0.5 MPa. The cell is placed in a solution with a water potential (ψw) of -0.4 MPa. What is the initial water potential of the cell and in which direction will net water movement occur?

  1. Cell ψw = -0.3 MPa; water moves into the cell.
  2. Cell ψw = -0.3 MPa; water moves out of the cell. (correct answer)
  3. Cell ψw = -1.3 MPa; water moves out of the cell.
  4. Cell ψw = 1.3 MPa; water moves into the cell.
Explanation: First, calculate the cell's water potential using the formula ψw = ψs + ψp. Cell ψw = (-0.8 MPa) + (+0.5 MPa) = -0.3 MPa. Water always moves down a water potential gradient, from a region of higher water potential to a region of lower water potential. The cell's potential is -0.3 MPa and the solution's is -0.4 MPa. Since -0.3 is a higher value (less negative) than -0.4, water will move from the cell to the surrounding solution.

Question 3

A root hair cell absorbs water from soil. The soil solution has a water potential (ψw) of -0.2 MPa. Which set of potentials for the root hair cell's cytoplasm would allow for the most rapid net influx of water?

  1. ψs = -0.2 MPa, ψp = 0 MPa
  2. ψs = -0.5 MPa, ψp = +0.2 MPa
  3. ψs = -0.3 MPa, ψp = +0.2 MPa
  4. ψs = -0.8 MPa, ψp = +0.1 MPa (correct answer)
Explanation: The rate of water movement is proportional to the water potential gradient (Δψw) between the soil and the cell. Water will move into the cell if the cell's ψw is lower (more negative) than the soil's ψw (-0.2 MPa). We calculate ψw for each option: A) ψw = -0.2 MPa (no gradient); B) ψw = -0.3 MPa (gradient = 0.1 MPa); C) ψw = -0.1 MPa (water would move out); D) ψw = -0.7 MPa (gradient = 0.5 MPa). The largest gradient is in option D, which would result in the most rapid water influx.

Question 4

A turgid plant cell is at equilibrium with a surrounding solution that has a solute potential (ψs) of -0.6 MPa. If the cell's own solute potential is -1.0 MPa, what is the pressure potential (ψp) of the cell?

  1. -1.6 MPa
  2. -0.4 MPa
  3. +0.4 MPa (correct answer)
  4. +1.6 MPa
Explanation: At equilibrium, the water potential of the cell (ψw_cell) must equal the water potential of the solution (ψw_solution). The solution is at atmospheric pressure, so its ψp = 0, meaning ψw_solution = ψs_solution = -0.6 MPa. For the cell, ψw_cell = ψs_cell + ψp_cell. Setting them equal: -0.6 MPa = -1.0 MPa + ψp_cell. Solving for ψp_cell gives: ψp_cell = -0.6 MPa + 1.0 MPa = +0.4 MPa.

Question 5

Cell X has a solute potential of -0.7 MPa and a pressure potential of +0.5 MPa. It is adjacent to Cell Y, which has a solute potential of -0.8 MPa and a pressure potential of +0.4 MPa. What will be the net direction of water movement between them and why?

  1. From Cell X to Cell Y, because Cell X has a higher water potential. (correct answer)
  2. From Cell Y to Cell X, because Cell Y has a more negative solute potential.
  3. From Cell Y to Cell X, because Cell Y has a lower pressure potential.
  4. There will be no net movement, because the difference in their potentials is minimal.
Explanation: Water moves down a water potential gradient. First, calculate the water potential (ψw) for each cell. For Cell X: ψw = -0.7 + 0.5 = -0.2 MPa. For Cell Y: ψw = -0.8 + 0.4 = -0.4 MPa. Since -0.2 MPa is a higher water potential than -0.4 MPa, water will move from the area of higher potential (Cell X) to the area of lower potential (Cell Y).

Question 6

In an experiment, potato cylinders are placed in sucrose solutions of varying concentrations. It is observed that there is no net change in mass in the 0.3 M sucrose solution. What can be concluded about the potato cells?

  1. The water potential of the potato cells is 0 MPa.
  2. The solute potential of the potato cells is equal to the solute potential of the 0.3 M sucrose solution. (correct answer)
  3. The pressure potential of the potato cells is equal to the solute potential of the 0.3 M sucrose solution.
  4. The potato cells have the same concentration of sucrose as the external solution.
Explanation: No change in mass indicates no net water movement. This occurs when the water potential inside the cells equals the water potential of the external solution. At this equilibrium point in an open solution, the cells are flaccid, meaning their pressure potential (ψp) is zero. Therefore, the cell's water potential (ψw) is equal to its solute potential (ψs). The solution's water potential is also equal to its solute potential. Thus, the solute potential of the cells is equal to the solute potential of the isotonic solution.

Question 7

A plant cell with a solute potential (ψs) of -1.2 MPa and a pressure potential (ψp) of +0.5 MPa is placed in a solution with a solute potential of -0.4 MPa. What is the water potential gradient (Δψw) between the solution and the cell, and which way will water move?

  1. Gradient = 0.3 MPa; water moves into the cell (correct answer)
  2. Gradient = 0.3 MPa; water moves out of the cell
  3. Gradient = 1.1 MPa; water moves out of the cell
  4. Gradient = 0.8 MPa; water moves into the cell
Explanation: First, calculate the water potential of the cell: ψw(cell) = ψs + ψp = -1.2 MPa + 0.5 MPa = -0.7 MPa. The solution has ψp = 0, so ψw(solution) = -0.4 MPa + 0 = -0.4 MPa. The gradient is |-0.4 - (-0.7)| = 0.3 MPa. Water moves from higher to lower potential: from solution (-0.4 MPa) to cell (-0.7 MPa), so water moves into the cell.

Question 8

A flaccid plant cell (ψp = 0) has a solute potential (ψs) of -1.1 MPa. It is placed in pure water (ψw = 0 MPa). Assuming the cell wall allows the cell to withstand the pressure, what will be the pressure potential (ψp) of the cell when it reaches equilibrium?

  1. -1.1 MPa
  2. 0 MPa
  3. +0.55 MPa
  4. +1.1 MPa (correct answer)
Explanation: The cell is placed in pure water (ψw = 0 MPa). Water will move into the cell down the water potential gradient, as the cell's initial ψw is -1.1 MPa. As water enters, turgor pressure builds up, and the pressure potential (ψp) increases from zero. Equilibrium is reached when the cell's water potential equals the water potential of the pure water. So, at equilibrium, ψw_cell = 0 MPa. Using the formula ψw = ψs + ψp, we get 0 MPa = -1.1 MPa + ψp. Solving for ψp gives ψp = +1.1 MPa.

Question 9

The opening of stomata is triggered by the active transport of K⁺ ions into guard cells. How does this initial event lead to the influx of water?

  1. The K⁺ ions increase the pressure potential (ψp) of the guard cells, drawing water in.
  2. The influx of K⁺ makes the solute potential (ψs) of the guard cells more negative, lowering their water potential (ψw). (correct answer)
  3. The K⁺ ions bind to water molecules, physically pulling them into the guard cells.
  4. The influx of K⁺ makes the solute potential (ψs) of the guard cells less negative, raising their water potential (ψw).
Explanation: Active transport increases the concentration of K⁺ ions (solutes) inside the guard cells. An increased solute concentration makes the solute potential (ψs) more negative. This, in turn, lowers the overall water potential (ψw = ψs + ψp) of the guard cells relative to the surrounding cells. This lower water potential creates a gradient that causes water to move into the guard cells via osmosis, increasing their turgor and opening the stoma.

Question 10

During a severe drought, the water potential of the soil can drop to -2.0 MPa. A plant's root cells have a solute potential of -1.4 MPa. What must be true for the plant to avoid losing water to the soil?

  1. The root cells must maintain a pressure potential of at least +0.6 MPa.
  2. The root cells' water potential must be lower than -2.0 MPa. (correct answer)
  3. The root cells must decrease their solute concentration to raise their solute potential.
  4. The plant must close its stomata, which directly increases the water potential of the roots.
Explanation: Water moves from a higher to a lower water potential. To avoid losing water to the soil, or to absorb water from it, the plant's root cells must have a water potential that is equal to or lower than the soil's water potential. Therefore, the root cells' ψw must be ≤ -2.0 MPa. While closing stomata helps conserve water, it does not directly change the root potential enough to overcome such a large soil water deficit. The only way to survive is to lower the internal water potential below that of the soil.

Question 11

A plant cell has a solute potential of -700 kPa and is in equilibrium with an external solution. If the cell's pressure potential is 250 kPa, what is the water potential of the external solution in megapascals (MPa)?

  1. -0.95 MPa
  2. -450 MPa
  3. +0.45 MPa
  4. -0.45 MPa (correct answer)
Explanation: First, calculate the cell's water potential in kilopascals (kPa): ψw = ψs + ψp = -700 kPa + 250 kPa = -450 kPa. Because the cell is at equilibrium with the solution, the solution must have the same water potential, -450 kPa. Second, convert this value to megapascals (MPa). Since 1 MPa = 1000 kPa, you divide by 1000: -450 kPa / 1000 = -0.45 MPa.

Question 12

If the pressure potential (ψp) of a plant cell is measured to be equal in magnitude but opposite in sign to its solute potential (ψs), what can be deduced about the cell's condition?

  1. The cell is flaccid and is likely in an isotonic solution.
  2. The cell is plasmolyzed and has lost all its water.
  3. The cell is fully turgid and is in equilibrium with pure water. (correct answer)
  4. The cell has a water potential equal to its solute potential.
Explanation: The condition is ψp = -ψs. The water potential of the cell is ψw = ψs + ψp. Substituting the condition into the equation gives ψw = ψs + (-ψs) = 0 MPa. A cell with a water potential of 0 MPa must be in equilibrium with an environment that also has a water potential of 0 MPa. This environment is pure water. A plant cell in pure water will take up water until it is fully turgid, at which point its internal ψw equals the external ψw.

Question 13

The solute potential (ψs) of a solution can be calculated using the formula ψs = -iCRT, where 'i' is the ionization constant, 'C' is the molar concentration, 'R' is the pressure constant, and 'T' is the absolute temperature. How would dissolving sucrose (i=1) versus sodium chloride (i≈1.8) at the same molar concentration affect the water potential of a solution?

  1. The sucrose solution would have a more negative water potential than the NaCl solution.
  2. The NaCl solution would have a positive water potential while the sucrose solution would be negative.
  3. Both solutions would have identical water potentials since their molar concentrations are the same.
  4. The NaCl solution would have a more negative water potential than the sucrose solution. (correct answer)
Explanation: The ionization constant 'i' reflects the number of particles a solute dissociates into in solution. Sucrose does not ionize (i=1), while NaCl dissociates into Na⁺ and Cl⁻ ions (i approaches 2, here given as 1.8). According to the formula ψs = -iCRT, a larger 'i' value results in a more negative solute potential for the same molar concentration 'C'. Since the water potential of an open solution is equal to its solute potential, the NaCl solution will have a more negative water potential than the sucrose solution.

Question 14

Aquaporins facilitate rapid osmosis across membranes. How would the presence of a high density of aquaporins affect the final water potential (ψw) and pressure potential (ψp) of a plant cell at equilibrium in a hypotonic solution, compared to a cell with few aquaporins?

  1. The final ψw and ψp would be significantly higher due to more efficient water transport.
  2. The final ψw and ψp would be the same, but equilibrium would be reached more quickly. (correct answer)
  3. The final ψw would be lower because aquaporins allow some solute leakage from the cell.
  4. The final ψp would be lower because the rapid influx of water damages the cell wall's integrity.
Explanation: Aquaporins are protein channels that affect the rate of water movement (i.e., membrane permeability), not the thermodynamic driving force. The final equilibrium state is determined by the initial water potential gradient and the physical properties of the cell (solute concentration and cell wall elasticity). Aquaporins allow the cell to reach this equilibrium state much faster, but they do not alter the final values of ψw or ψp at which equilibrium occurs.

Question 15

Consider three adjacent plant cells: P, Q, and R. Their potentials are given in MPa. Cell P: ψs = -0.9, ψp = +0.6. Cell Q: ψs = -1.0, ψp = +0.5. Cell R: ψs = -0.8, ψp = +0.2. What will be the net direction of water movement between these cells?

  1. From R to Q, and from Q to P.
  2. From P to R, and from Q to R.
  3. From P to Q, and from Q to R. (correct answer)
  4. From R to P, and from Q to P.
Explanation: To determine the direction of water movement, calculate the water potential (ψw) for each cell: ψw = ψs + ψp. For P: ψw = -0.9 + 0.6 = -0.3 MPa. For Q: ψw = -1.0 + 0.5 = -0.5 MPa. For R: ψw = -0.8 + 0.2 = -0.6 MPa. Water moves from higher to lower water potential. The potentials in descending order are P (-0.3 MPa) > Q (-0.5 MPa) > R (-0.6 MPa). Therefore, water will move from P to Q, and from Q to R.

Question 16

A plant cell is at the point of incipient plasmolysis when placed in a particular sucrose solution. At this stage, the solute potential (ψs) of the cell is determined to be -0.9 MPa. What are the approximate values for the cell's pressure potential (ψp) and water potential (ψw)?

  1. ψp = +0.9 MPa; ψw = 0 MPa
  2. ψp = 0 MPa; ψw = -0.9 MPa (correct answer)
  3. ψp = -0.9 MPa; ψw = -1.8 MPa
  4. ψp = 0 MPa; ψw = +0.9 MPa
Explanation: Incipient plasmolysis is the point where the cell membrane just begins to pull away from the cell wall. At this point, the cell is flaccid, meaning it is exerting no pressure on the cell wall. Therefore, the pressure potential (ψp) is zero. The water potential of the cell is calculated as ψw = ψs + ψp. With ψp = 0 MPa, the cell's water potential is equal to its solute potential: ψw = -0.9 MPa + 0 MPa = -0.9 MPa.

Question 17

A herbaceous plant wilts on a hot day. Which statement correctly describes the changes in water potential components within its leaf parenchyma cells?

  1. Solute potential (ψs) becomes less negative due to water loss, causing water potential (ψw) to rise.
  2. Pressure potential (ψp) increases significantly as the cells attempt to conserve water.
  3. Pressure potential (ψp) decreases towards zero, causing the overall water potential (ψw) to become more negative. (correct answer)
  4. Both solute potential (ψs) and pressure potential (ψp) approach zero, resulting in a water potential (ψw) of zero.
Explanation: Wilting is the loss of turgidity due to excessive water loss. This means the pressure exerted by the cell contents on the cell wall decreases. Therefore, the pressure potential (ψp), also known as turgor pressure, decreases and approaches zero. Since ψw = ψs + ψp, a decrease in ψp will make the overall water potential (ψw) more negative (as ψs is already negative), promoting water uptake if available.

Question 18

Which statement best explains the significance of the water potential (ψw) of pure water being defined as 0 MPa at standard atmospheric pressure?

  1. It signifies that pure water has no potential to do work and is inert.
  2. It provides a universal reference point against which the water potential of all solutions and biological tissues is compared. (correct answer)
  3. It indicates that the solute potential and pressure potential of pure water are both zero.
  4. It reflects the equilibrium point where water movement ceases under all conditions.
Explanation: By convention, the water potential of pure water under standard conditions is set to zero. This value serves as a fundamental reference point. The addition of solutes makes water potential negative, and the application of positive pressure makes it positive. This allows for a standardized, quantitative comparison of the tendency of water to move from one area to another.

Question 19

Water moves up the xylem from roots to leaves, driven by transpiration. How does the water potential (ψw) in the leaf xylem typically compare to the ψw in the root xylem?

  1. Leaf xylem ψw is higher due to the positive pressure generated by transpiration.
  2. Leaf xylem ψw is significantly lower, primarily due to a large negative pressure potential (tension). (correct answer)
  3. Leaf xylem ψw is slightly lower, primarily due to a more negative solute potential from mineral accumulation.
  4. Leaf xylem ψw is approximately the same as root xylem ψw to maintain a continuous column of water.
Explanation: According to the cohesion-tension theory, the evaporation of water from leaf surfaces (transpiration) creates a negative pressure, or tension, in the xylem. This negative pressure potential (ψp) makes the overall water potential (ψw) in the leaf xylem extremely low (very negative). This creates a steep water potential gradient between the roots (higher ψw) and the leaves (lower ψw), which pulls the water column upwards.

Question 20

A red blood cell, which lacks a cell wall, has a cytoplasm with a water potential of -0.7 MPa. It is placed in a solution with a water potential of -0.2 MPa. What is the most likely outcome?

  1. The cell will shrink (crenate) as water moves out down its concentration gradient.
  2. The cell will swell and may burst (undergo lysis) due to the net influx of water. (correct answer)
  3. The cell will remain unchanged as the cell membrane prevents water movement.
  4. The cell will become turgid as an internal pressure balances the water potential gradient.
Explanation: Water moves from a region of higher water potential to one of lower water potential. The solution (-0.2 MPa) has a higher water potential than the cell's cytoplasm (-0.7 MPa). Therefore, water will move into the red blood cell. Because animal cells lack a rigid cell wall, they cannot build up significant internal pressure (turgor) to counteract this influx. The cell will continue to swell and is likely to burst (undergo cytolysis).