3.1.6Exchange and Transport
Gas Exchange vs Water Loss — AQA A Level Biology 7402 Revision Notes & Topical Questions
5 min read 9 parts Lesson 9 of 48 1 quiz
3.1.6 Gas Exchange vs Water Loss — AQA A Level Biology 7402 notes: Explain the conflict between gas exchange and water loss and describe the adaptations of insects and xerophytic plants that reduce water loss.Efficient gas exchange needs a large, thin, moist surface in contact with the air.
- evaporation
- waterproof exoskeleton
- transpiration
- waxy cuticle
- Xerophytes
- Exam-style question 1
Section 1 of 9The problem
Efficient gas exchange needs a large, thin, moist surface in contact with the air. Such a surface is also ideal for evaporation of water. Terrestrial organisms therefore face a conflict: the features that favour gas exchange also lead to dehydration. Organisms solve this by compromise, having mechanisms to reduce water loss and to open exchange surfaces only when needed.
Section 2 of 9Insects
Insects reduce water loss in several ways:
- and which are opened only when oxygen is needed or carbon dioxide has built up;
- they may open and close in cycles;
The trade-off is that when spiracles are open for gas exchange, water vapour is lost. Insects that are very active therefore lose more water.
Section 3 of 9Plants: stomata and the leaf
In a leaf, water vapour is lost by transpiration through open stomata.
- when they lose water, they become flaccid and the stoma closes.
- and close in the dark or when the plant is short of water.
- (abscisic acid is the hormone that triggers closure in drought)
Section 4 of 9Xerophytes
Xerophytes are plants adapted to arid conditions, such as marram grass, cacti and some conifers. Their adaptations reduce transpiration but still allow gas exchange.
| Adaptation | How it reduces water loss |
|---|---|
| Thick waxy cuticle | Increases the diffusion path and is waterproof |
| Rolled leaves (marram grass) | Trap a region of humid air so the water potential gradient to the outside is reduced |
| Hairs on the leaf surface | Trap moist air close to the surface, reducing the gradient |
| Sunken stomata (in pits) | Shelter stomata from air movement so humid air is retained |
| Reduced leaf surface area (spines, needles) | Fewer stomata and a smaller surface for evaporation |
| Fewer stomata | Reduces the number of pores for water vapour to escape |
| Succulent tissues, deep or wide roots | Store water or access it |
In each adaptation, the principle is to reduce the water potential gradient between the leaf and the air, to increase the diffusion path, or to reduce the surface area. In an exam, state the feature and its effect in terms of diffusion of water vapour.
Section 5 of 9Investigating transpiration
A potometer measures the rate of water uptake by a leafy shoot, which approximates transpiration.
Section 6 of 9Evaluating adaptations in desert and aquatic environments
For a data-based question, look at how a species of plant from a dry habitat compares with one from a wet habitat: the xerophyte may have fewer stomata per square millimetre, thicker cuticle and more hairs, and its rate of transpiration will be lower under the same conditions.
- and a hydrophyte,
- such as a water lily, has stomata on the upper surface only,
- since the lower surface is in contact with water,
- and large air spaces to help it float and distribute oxygen.
- so growth of xerophytes is slower,
- which is the cost of living in arid environments.
Section 7 of 9Exam-style questions
Explain why insects lose water when they open their spiracles, and how they reduce this loss. [4]4 marks
11. Open spiracles allow water vapour to diffuse out down the water potential gradient.22. Spiracles are closed when gas exchange is not needed.33. The exoskeleton has a waxy waterproof cuticle.44. Hairs around spiracles trap humid air to reduce the gradient.Explain how the structure of marram grass leaves reduces water loss. [4]4 marks
11. Leaves are rolled, trapping moist air so that the water potential gradient is reduced.22. Stomata are in pits, which shelter them from air movement.33. Hairs trap moist air.44. A thick cuticle increases the diffusion path for water vapour.A potometer bubble moved 36 mm in 12 minutes in a capillary tube of radius 0.5 mm. Calculate the rate of water uptake in mm3 per minute. [3]3 marks
1. Cross-sectional area = π x 0.52 = 0.785 mm2.
2. Volume = 0.785 x 36 = 28.3 mm3.
3. Rate = 28.3 / 12 = 2.36 mm3 per minute.
Section 8 of 9Key terms
| Term | Meaning |
|---|---|
| Transpiration | Evaporation of water from the leaves through the stomata |
| Xerophyte | Plant adapted to live in dry conditions |
| Cuticle | Waxy waterproof layer covering the epidermis |
| Guard cell | Cell that controls the opening of a stoma |
| Water potential gradient | Difference in water potential driving the diffusion of water vapour |
| Potometer | Apparatus for measuring the rate of water uptake by a shoot |
| Spiracle | Pore in the insect cuticle that can be closed to reduce water loss |
Section 9 of 9Exam technique
Marked practiceQuizzes for this lesson
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