HSCScienceExam practice
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Biology  ·  Year 11  ·  Module 2  ·  Lesson 17

HSC Exam Practice

Transpiration: Factors and Measurement

8 questions / 3 sections / 26 marks total
Section 1

Short answer

1.Short answer

1.1

Define transpiration and explain why a potometer does not measure it directly.

3marks Band 3
1.2

Explain why high humidity reduces transpiration rate. In your answer, refer to the water potential gradient.

3marks Band 3–4
1.3

Explain how a xerophyte with sunken stomata reduces its rate of transpiration. In your answer, name the mechanism and explain how it reduces the driving force for water vapour loss.

3marks Band 4
1.4

Outline the mechanism by which light intensity affects stomatal aperture. Start from light absorption by guard cells and end with a change in pore width.

3marks Band 4
Section 2

Data response

2.Data response, potometer results under three conditions

2.1

A student uses a potometer to compare transpiration rates under three conditions. All other variables are held constant. The table below shows the mean distance (mm) the air bubble moved in 10 minutes across three trials.

0 10 20 30 40 50 Mean bubble movement (mm / 10 min) Control 13 Fan (wind) 24 Low humidity 32 Experimental condition
Figure 2.1. Mean bubble movement (mm) in 10 minutes for three potometer conditions. Control: still air, 22°C, 60% humidity, dim light. Fan: fan added to control. Low humidity: 30% humidity, otherwise as control. Hypothetical data.

(a) Describe the trend shown in the data, quoting at least two values. 2 marks

(b) Account for the difference in bubble movement between the control and the fan condition. Refer to the boundary layer and water potential gradient in your explanation. 3 marks

(c) Identify one limitation of using a potometer in this investigation and suggest an improvement. 2 marks

7marks Band 3–5
Section 3

Extended response

3.Extended response

3.1

A student claims: “On a hot, dry, windy day a plant will wilt faster because transpiration rate is much higher than on a cool, humid, still day.” Assess this claim by explaining the effect of each named factor on transpiration rate and describing how the combined effect leads to wilting. In your response, refer to the water potential gradient and guard cell function where relevant.

7marks Band 4–6

Biology · Year 11 · Module 2 · Lesson 17

Answer Key & Marking Guidelines

1.1

Section 1 · Short answer · 3 marks · Band 3

Sample response. Transpiration is the evaporation of water from mesophyll cells into leaf air spaces and its diffusion through open stomata into the external atmosphere. A potometer measures water uptake by the cut shoot, not transpiration directly. Water uptake includes transpiration but also water incorporated into photosynthesis products and water retained in growing cells, therefore the potometer slightly overestimates the transpiration component.

Marking notes. 1 mark for a correct definition of transpiration (evaporation from mesophyll + diffusion through stomata); 1 mark for identifying that the potometer measures water uptake, not transpiration; 1 mark for explaining at least one reason uptake differs from transpiration (photosynthesis / cell growth / metabolic use).

1.2

Section 1 · Short answer · 3 marks · Band 3–4

Sample response. Transpiration is driven by the difference in water vapour concentration (water potential) between the nearly saturated leaf air spaces (~99% relative humidity) and the outside air. High humidity means the outside air already contains a high concentration of water vapour, so the water potential difference between the leaf interior and the atmosphere is small. Because the driving gradient is reduced, water vapour diffuses out through the stomata more slowly. At 100% humidity, there is no gradient and no net transpiration occurs.

Marking notes. 1 mark for identifying that transpiration is driven by a water potential/concentration gradient between leaf interior and atmosphere; 1 mark for explaining that high humidity reduces this gradient (outside air is closer to saturation); 1 mark for conclusion that smaller gradient means slower diffusion and lower transpiration rate.

1.3

Section 1 · Short answer · 3 marks · Band 4

Sample response. Sunken stomata are recessed into crypts or pits below the leaf surface. Water vapour transpired through the stomata accumulates within the enclosed crypt, creating a humid still-air boundary layer immediately outside the stomatal pore. This boundary layer has a higher water vapour concentration than the bulk atmosphere outside. The driving force for diffusion is the water potential difference between the leaf interior and the air immediately outside the pore, but with the boundary layer in place, this difference is much smaller than the difference between the leaf and the dry bulk air. Diffusion therefore slows, reducing transpiration rate.

Marking notes. 1 mark for describing that stomata are recessed into crypts/pits so transpired water vapour accumulates within the crypt; 1 mark for naming the mechanism: boundary layer / humid still-air layer in the crypt; 1 mark for explaining how this reduces the water potential gradient (comparison is between leaf and humid crypt air, not between leaf and dry bulk air), reducing the driving force for diffusion.

1.4

Section 1 · Short answer · 3 marks · Band 4

Sample response. When light is absorbed by guard cells, they photosynthesise, producing ATP. This ATP powers active transport of K¹+ ions into the guard cells. The increase in intracellular K¹+ lowers the water potential inside the guard cells. Water then enters the guard cells by osmosis from surrounding cells, increasing their turgor pressure. The turgid guard cells bow outward due to their elongated curved shape and the uneven thickening of their cell wall, widening the stomatal pore.

Marking notes. 1 mark for light → photosynthesis → ATP → active K¹+ accumulation in guard cells; 1 mark for K¹+ accumulation → lower water potential → osmotic water entry → increased turgor; 1 mark for increased turgor → guard cells bow outward → stomatal pore opens/widens.

2.1(a)

Section 2 · Data response · 2 marks · Band 3

Sample response. Bubble movement increases across all three conditions, with control (13 mm), fan/wind (24 mm), and low humidity (32 mm) showing a progressive increase. The low humidity condition produced the greatest bubble movement, approximately 2.5 times greater than the control.

Marking notes. 1 mark for identifying the overall trend (water uptake increases from control to fan to low humidity); 1 mark for quoting at least two specific values from the graph to support the description.

2.1(b)

Section 2 · Data response · 3 marks · Band 4–5

Sample response. In the control (still air), transpired water vapour accumulates immediately outside the stomata as a humid boundary layer. This partially saturates the air at the stomatal pore, reducing the effective water potential gradient between the leaf interior and the air at the pore. In the fan condition, the wind constantly removes this boundary layer, replacing the humid stagnant air around the stomata with drier bulk air. The water potential gradient at the stomatal pore is therefore steeper, driving faster diffusion of water vapour out of the leaf. This accelerated water vapour loss is reflected in greater water uptake by the shoot (24 vs 13 mm per 10 minutes).

Marking notes. 1 mark for identifying the boundary layer in still air (humid air accumulates outside stomata); 1 mark for explaining that wind removes the boundary layer and replaces it with drier bulk air; 1 mark for connecting boundary layer removal to a steeper water potential gradient → faster diffusion of water vapour → greater water uptake.

2.1(c)

Section 2 · Data response · 2 marks · Band 4

Sample response. Limitation: the potometer measures water uptake, not transpiration directly, a small amount of water taken up is used in photosynthesis rather than being transpired, so results slightly overestimate transpiration rate. Improvement: compare potometer results with a gravimetric method (weighing the shoot before and after the trial) to determine the actual amount of water lost from the leaves, providing a direct measure of transpiration. Accept also: limitation = using the same shoot across all three conditions (shoot may deteriorate or acclimate); improvement = use a fresh shoot for each condition re-cut under water.

Marking notes. 1 mark for a valid, specific limitation (water uptake vs transpiration; same shoot deterioration; no root system; insufficient repeats); 1 mark for a corresponding valid improvement that addresses the stated limitation.

3.1

Section 3 · Extended response · 7 marks · Band 4–6

Sample response. The student’s claim is correct: a hot, dry, windy day produces much higher transpiration than a cool, humid, still day because all three named factors independently increase transpiration rate, and their effects combine.

High temperature increases the kinetic energy of water molecules in the leaf mesophyll, accelerating evaporation from cell walls into the leaf air spaces. Warm air also has greater capacity to hold water vapour, so the atmosphere is further from saturation at any given absolute humidity. Both effects steepen the water potential gradient between the humid leaf interior (~99% relative humidity) and the outside air, driving faster diffusion of water vapour through stomata.

Dry air (low humidity) means the outside atmosphere has low water vapour concentration. The water potential difference between the near-saturated leaf air spaces and the dry bulk atmosphere is very large, providing a strong driving force for diffusion. On a cool, humid day, this gradient is small.

Wind removes the humid boundary layer that accumulates just outside the stomata in still conditions. By constantly replacing this humid air with dry bulk air, wind maintains the maximum possible water potential gradient at the stomatal pore throughout the day. Guard cells are not directly affected by moderate wind, stomata remain open, so the widened pathway and the steep gradient both contribute to a high rate.

Together, the three factors push transpiration rate far above the rate at which roots can absorb water from soil. As water leaves the plant faster than it is replaced, cells throughout the leaf and stem lose water to the xylem by osmosis, their turgor pressure falls, and without turgor the cells become flaccid. Guard cells also lose turgor and the stomata begin to close as a protective response, but if soil water is unavailable or the root system cannot supply water fast enough, wilting occurs as the whole plant loses the hydraulic pressure that keeps it upright.

Marking criteria.

  • 1 markCorrectly states the claim is broadly correct and that the three factors combine to increase transpiration.
  • 1 markExplains the effect of high temperature on transpiration, explicitly connecting it to the water potential gradient (kinetic energy + atmospheric holding capacity).
  • 1 markExplains the effect of dry air/low humidity on the water potential gradient (large difference between near-saturated leaf and dry atmosphere → fast diffusion).
  • 1 markExplains the effect of wind on the boundary layer and how this maintains a steep gradient at the stomatal pore.
  • 1 markLinks the combined high transpiration rate to water deficit in the plant: water leaves faster than root uptake can supply.
  • 1 markExplains wilting mechanistically: loss of cell water → reduced turgor pressure → cells become flaccid → plant wilts.
  • 1 markQuality of scientific language: uses precise terms consistently throughout (water potential gradient, stomatal aperture, turgor pressure, boundary layer, diffusion) and presents a coherent causal chain from environmental conditions to cellular consequence.