Revision Library

revisionlibrary.co.uk

GCSE · Biology

B2c Plant Tissues, Transport and Transpiration

AQA 8464 · Calculator allowed · about 100 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Key Facts: Plant Tissues, Transport and Transpiration

Original text written for Revision Library.

Plants are organised into tissues and organs that transport water, mineral ions and dissolved food around the plant. A leaf is an organ made of several tissues: the upper epidermis is covered by a waxy cuticle that reduces water loss; the palisade mesophyll, packed with chloroplasts, lies just beneath it to absorb the maximum amount of light for photosynthesis; the spongy mesophyll below has large air spaces that allow gases to diffuse to and from the mesophyll cells; and the lower epidermis contains most of the stomata, tiny pores surrounded by a pair of guard cells that open and close to control gas exchange and water loss. Xylem tissue is made of dead, hollow cells strengthened with lignin, joined end to end into continuous tubes that transport water and mineral ions from the roots to the leaves in one direction only. Phloem tissue is made of living cells and transports dissolved sugars, such as sucrose, and amino acids around the plant in either direction, from sources (where they are made or stored) to sinks (where they are used or stored); this process is called translocation. Transpiration is the loss of water vapour from a plant, mainly through the stomata, and it is the driving force that pulls a continuous stream of water and mineral ions up through the xylem, called the transpiration stream. The rate of transpiration is affected by light intensity, temperature, humidity and air movement, and can be measured in the laboratory using a potometer.

The list below gives descriptions of five plant structures.

Descriptions:
1. A dead, hollow, lignified cell that transports water and mineral ions.
2. A living, elongated cell that transports dissolved sugars and amino acids around the plant.
3. A tightly packed cell containing many chloroplasts, found near the upper surface of a leaf.
4. A small pore, usually in the lower epidermis of a leaf, that allows gases to diffuse in and out.
5. A specialised cell that changes shape to open and close a pore in the leaf surface.

For each term below, write the number (1-5) of the description that matches it.

(a)Xylem cell(1)

(b)Phloem cell(1)

(c)Palisade mesophyll cell(1)

(d)Stoma(1)

(e)Guard cell(1)

(Total for Question 1 is 5 marks)

The upper epidermis of a leaf is covered by a waxy cuticle.

(Total for Question 2 is 2 marks)

Xylem tissue is adapted for transporting water from the roots to the leaves.

(Total for Question 3 is 2 marks)

Translocation is the movement of substances through phloem tissue.

(a)Name the tissue responsible for translocation.(1)

(b)Name two substances transported by this tissue.(2)

(Total for Question 4 is 3 marks)

Xylem and phloem both transport substances around a plant, but in different ways.

(Total for Question 5 is 2 marks)

A simplified diagram shows a cross-section through a leaf. From top to bottom, the layers are labelled: P (a single layer of transparent cells covered by a thick waxy layer), Q (a layer of tall, tightly packed cells directly beneath P), R (a layer of loosely packed, irregularly shaped cells with large air spaces between them), S (a single layer of cells containing two bean-shaped cells with a small gap between them, on the lower surface), and T (a bundle of specialised transport cells running through the middle of the leaf).

Figure (to be drawn): Cross-section of a leaf drawn as horizontal bands from top to bottom: P = upper epidermis with waxy cuticle on top; Q = palisade mesophyll (tall, tightly packed cells with many small dots representing chloroplasts, concentrated near the top of each cell); R = spongy mesophyll (loosely packed, irregular cells with large white gaps representing air spaces); S = lower epidermis with a pair of bean-shaped guard cells enclosing a small central gap (a stoma); T = a vascular bundle (a cluster of small tube-shaped cells) running horizontally through layers Q and R.

(a)Identify layers P, Q, R, S and T.(5)

(b)Explain why layer Q contains many chloroplasts positioned near the upper surface of the leaf.(2)

(c)Suggest why layer R has large air spaces between its cells.(2)

(Total for Question 6 is 9 marks)

Stomata are small pores in the leaf surface, each surrounded by a pair of guard cells.

(a)Name the cells that surround and control the size of a stoma.(1)

(b)Describe how guard cells cause a stoma to open.(3)

(c)Explain why guard cells often close stomata during a water shortage.(2)

(Total for Question 7 is 6 marks)

Root hair cells are adapted for the efficient uptake of water from the soil.

(a)Name the process by which water moves from the soil into root hair cells.(1)

(b)State two structural adaptations of root hair cells and, for each one, explain how it increases the rate of water uptake.(4)

(Total for Question 8 is 5 marks)

Root hair cells use active transport to absorb mineral ions from the soil, even when the concentration of ions in the soil is lower than inside the cell. Explain how this active transport of mineral ions can affect the movement of water into the cell by osmosis.

(Total for Question 9 is 3 marks)

Water travels from the roots to the leaves of a plant, driven by transpiration.

(a)Define transpiration.(1)

(b)Describe the pathway taken by water from being absorbed by the roots to being lost from the leaves.(4)

(c)Explain how transpiration helps to supply the plant with water and dissolved mineral ions.(2)

(Total for Question 10 is 7 marks)

A student investigated the effect of air temperature on the rate of water uptake by a leafy shoot, using a potometer. Light intensity, humidity and air movement were kept constant throughout.

Temperature (deg C)	Rate of water uptake (mm³/min)
10	12
20	24
30	39
40	41

(a)Calculate the percentage increase in the rate of water uptake between 10 deg C and 20 deg C. Show your working.(2)

(b)Describe the overall trend shown by the data.(2)

(c)Explain, in terms of particles, why increasing temperature increases the rate of transpiration.(3)

(d)Suggest one reason why the increase in rate slows between 30 deg C and 40 deg C.(2)

(Total for Question 11 is 9 marks)

A student set up a potometer to investigate the rate of water uptake by a leafy shoot, as shown in the diagram.

Figure (to be drawn): A leafy shoot is fitted through a rubber bung into the top of a horizontal capillary tube, which is completely filled with water and connected to a graduated scale. A side reservoir with a tap allows an air bubble to be reintroduced into the tube. Petroleum jelly (Vaseline) is smeared around the bung where the shoot enters the apparatus.

(a)Explain why the shoot should be cut underwater and at a slant before being inserted into the potometer.(2)

(b)Explain why all joints in the apparatus, such as around the bung, must be sealed with petroleum jelly.(1)

(c)A student wants to investigate how humidity affects the rate of transpiration using this potometer. State a suitable independent variable, dependent variable, and one variable that should be controlled.(3)

(d)Explain why the rate of water uptake measured using a potometer is not exactly the same as the rate of transpiration.(2)

(Total for Question 12 is 8 marks)

Using the same potometer described in Question 12, the student measured how far an air bubble moved along the capillary tube in a fixed time, then repeated the experiment with a fan blowing air across the shoot.

(a)The capillary tube has a diameter of 2.0 mm. Calculate the cross-sectional area of the tube, using the equation: area = π x radius^2. Give your answer to 3 significant figures.(2)

(b)Without the fan, the air bubble moved 45 mm along the tube in 3.0 minutes. Calculate the volume of water taken up by the shoot, and hence the rate of water uptake in mm³/min. Give your final answer to 3 significant figures.(3)

(c)Convert the rate of water uptake calculated in part (b) into cm³ per hour.(2)

(d)With the fan switched on, the bubble moved 63 mm in the same 3.0 minutes. Calculate the percentage increase in the rate of water uptake caused by the fan, compared with your answer to part (b).(3)

(e)Use your answer to part (d) to explain, in terms of diffusion, why increasing air movement increases the rate of transpiration.(2)

(Total for Question 13 is 12 marks)

In summer, a broad bean plant transports sucrose made in its leaves down to its roots for storage. In the following spring, the same plant transports sucrose from its roots up to its new growing shoots.

(a)Name the tissue responsible for transporting the sucrose.(1)

(b)Using the example given, explain what is meant by a 'source' and a 'sink' during translocation.(2)

(c)Explain why it is an advantage that phloem can transport substances in either direction.(2)

(Total for Question 14 is 5 marks)

Water lilies are plants that float on the surface of ponds, with the upper surface of each leaf exposed to the air and the lower surface resting flat on the water. Unlike most land plants, water lilies have almost all of their stomata on the upper surface of the leaf. Suggest why.

(Total for Question 15 is 3 marks)

A student used a light microscope to count the number of stomata on the lower surface of a leaf. She viewed the leaf under a magnification at which the circular field of view had a diameter of 1.5 mm, and counted 6 stomata within it.

Figure (to be drawn): A circular microscope field of view, diameter 1.5 mm, with 6 small kidney-shaped stomata (each formed from a pair of guard cells) scattered across the circle.

(a)Calculate the area of the field of view, in mm^2. Give your answer to 3 significant figures.(2)

(b)Calculate the stomatal density, in stomata per mm^2.(2)

(c)The student repeated the count on the upper surface of the same leaf and found a much lower stomatal density. Suggest one advantage to the plant of having fewer stomata on the upper surface of the leaf.(2)

(d)Suggest and explain how the stomatal density of a leaf might differ between a plant adapted to a hot, dry habitat and a plant adapted to a cool, wet habitat.(3)

(Total for Question 16 is 9 marks)

A gardener notices that a potted plant, standing in a well-lit, warm greenhouse, wilts (droops) more quickly on windy days than on still days, even though she waters it the same amount each day. Explain, as fully as you can, why the plant loses water more quickly on windy days, and suggest what might happen to the plant's rate of photosynthesis as a result.

(Total for Question 17 is 6 marks)

Mark scheme · B2c Plant Tissues, Transport and Transpiration

Question 1

(a) B1 1
(a) Answer: 1
(b) B1 2
(b) Answer: 2
(c) B1 3
(c) Answer: 3
(d) B1 4
(d) Answer: 4
(e) B1 5
(e) Answer: 5

Question 2

B1 it acts as a waterproof layer, reducing water loss by evaporation directly through the epidermis, oe
B1 it is transparent, allowing light to pass through to the palisade mesophyll for photosynthesis, oe
Answer: The waxy cuticle reduces water loss by evaporation (it is waterproof), and it is transparent so light can pass through to the palisade mesophyll.

Question 3

B1 xylem cells are dead, and joined end to end with no end walls/cross walls, forming a continuous hollow tube through which water can flow freely, oe
B1 the cell walls are thickened/strengthened with lignin, giving support and preventing the tube from collapsing, oe
Answer: Xylem cells are dead and joined end to end into a continuous hollow tube, with walls thickened by lignin, so water can flow freely while the tube stays supported and does not collapse.

Question 4

(a) B1 phloem
(a) Answer: Phloem
(b) B1 dissolved sugars/sucrose
(b) B1 amino acids
(b) Answer: Dissolved sugars (sucrose) and amino acids.

Question 5

B1 xylem transports substances (water and mineral ions) in one direction only, from roots to leaves, oe
B1 phloem transports substances (sugars and amino acids) in either direction, from sources to sinks, oe
Answer: Xylem transport is one-way, from roots to leaves, whereas phloem transport can go in either direction, from a source to a sink.

Question 6

(a) B1 P: upper epidermis
(a) B1 Q: palisade mesophyll (layer)
(a) B1 R: spongy mesophyll (layer)
(a) B1 S: lower epidermis
(a) B1 T: vascular bundle/xylem and phloem
(a) Answer: P: upper epidermis. Q: palisade mesophyll. R: spongy mesophyll. S: lower epidermis. T: vascular bundle (xylem and phloem).
(b) B1 light enters the leaf from above, oe
(b) B1 so positioning many chloroplasts near the top allows them to absorb the maximum amount of light for photosynthesis before it is absorbed or scattered by deeper tissue, oe
(b) Answer: Light enters the leaf from above, so having many chloroplasts positioned near the top of the palisade layer means they can absorb the maximum amount of light for photosynthesis.
(c) B1 the air spaces allow gases (carbon dioxide, oxygen and water vapour) to diffuse freely to and from the mesophyll cells, oe
(c) B1 this increases the surface area available for gas exchange within the leaf, and connects the mesophyll cells to the stomata for gas exchange with the air outside, oe
(c) Answer: The air spaces allow gases such as carbon dioxide, oxygen and water vapour to diffuse freely between the mesophyll cells and the stomata, increasing the surface area available for gas exchange.

Question 7

(a) B1 guard cells
(a) Answer: Guard cells
(b) B1 when water is available, guard cells take up water by osmosis and become turgid, oe
(b) B1 the outer wall of each guard cell is thinner than the inner wall, so the outer wall stretches more as the cell swells, oe
(b) B1 this uneven stretching makes the pair of guard cells curve/bend apart, opening the pore between them, oe
(b) Answer: Guard cells take up water by osmosis and become turgid; because their outer wall is thinner than their inner wall, it stretches more, making the pair of guard cells curve apart and open the stoma.
(c) B1 closing the stomata reduces the diffusion of water vapour out of the leaf, reducing water loss by transpiration, oe
(c) B1 this helps to conserve the plant's remaining water and prevents excessive water loss/wilting, oe
(c) Answer: Closing the stomata reduces water vapour loss by transpiration, helping the plant conserve its remaining water and avoid wilting.

Question 8

(a) B1 osmosis
(a) Answer: Osmosis
(b) B1 adaptation 1: a long, thin extension (root hair) giving a large surface area (in contact with soil water), oe
(b) B1 explanation 1: a larger surface area allows more water to move into the cell by osmosis at the same time, oe
(b) B1 adaptation 2: a thin cell wall/membrane (any one further valid adaptation, e.g. large numbers of mitochondria), oe
(b) B1 explanation 2: a thin wall gives a short diffusion pathway/distance so water can move in more quickly (or: many mitochondria release energy by respiration for the active transport of mineral ions, which helps maintain the water potential gradient needed for osmosis), oe
(b) Answer: Root hair cells have a long, thin extension giving a large surface area, which allows more water to enter by osmosis at once; they also have a thin cell wall, giving a short diffusion pathway so water enters more quickly.

Question 9

B1 active transport uses energy (from respiration) to move mineral ions into the root hair cell against their concentration gradient, oe
B1 this increases/maintains the concentration of dissolved solutes inside the cell, lowering the water potential inside the cell relative to the water potential of the soil water outside, oe
B1 this larger water potential gradient increases the rate at which water moves into the cell by osmosis, oe
Answer: Active transport uses energy from respiration to move mineral ions into the root hair cell against their concentration gradient. This raises the solute concentration (lowers the water potential) inside the cell compared with the soil water outside, creating or maintaining a water potential gradient that increases the rate of osmosis of water into the cell.

Question 10

(a) B1 the loss/evaporation of water vapour from a plant, mainly through the stomata (in the leaves), oe
(a) Answer: Transpiration is the loss of water vapour from a plant, mainly through the stomata in the leaves.
(b) B1 water is absorbed from the soil into root hair cells by osmosis, oe
(b) B1 water passes across the root (from cell to cell) and enters the xylem, oe
(b) B1 water moves up through the xylem, from the root, through the stem, and into the leaves, oe
(b) B1 water moves out of the xylem into the mesophyll cells, where it evaporates into the air spaces and diffuses out of the leaf through open stomata, oe
(b) Answer: Water is absorbed into root hair cells by osmosis, passes across the root into the xylem, moves up the xylem through the stem to the leaves, then moves into the mesophyll cells where it evaporates into the leaf's air spaces and diffuses out through open stomata.
(c) B1 as water evaporates from the mesophyll cells and diffuses out of the stomata, it is replaced by water pulled up through the xylem from the roots, oe
(c) B1 this continuous pull (the transpiration stream) also carries dissolved mineral ions up through the xylem to where they are needed for growth, oe
(c) Answer: Water lost from the leaves by transpiration is replaced by water pulled up through the xylem from the roots (the transpiration stream), and this same flow carries dissolved mineral ions up to where they are needed.

Question 11

(a) M1 (24 - 12) / 12 x 100, oe
(a) A1 100% cao
(a) Answer: 100%
(b) B1 as temperature increases, the rate of water uptake increases overall, oe
(b) B1 the increase becomes smaller at higher temperatures, e.g. the rate rises by 15 mm³/min between 20 and 30 deg C but only 2 mm³/min between 30 and 40 deg C, so the rate begins to level off, oe with correct use of data
(b) Answer: The rate of water uptake increases as temperature increases, but the size of the increase gets smaller at higher temperatures, with the rate levelling off between 30 and 40 deg C.
(c) B1 water molecules gain (kinetic) energy and move faster as temperature increases, oe
(c) B1 this increases the rate of evaporation of water from the surfaces of the mesophyll cells into the air spaces of the leaf, oe
(c) B1 it also increases the difference in water vapour concentration between inside the leaf and the air outside, increasing the rate of diffusion of water vapour out through the stomata, oe
(c) Answer: Higher temperature gives water molecules more kinetic energy, so they evaporate faster from the mesophyll cells into the leaf's air spaces; this increases the water vapour concentration gradient between the leaf and the outside air, so water vapour diffuses out through the stomata faster.
(d) B1 at higher temperatures, the plant may begin to close its stomata (or guard cells may lose turgor), oe
(d) B1 to prevent excessive water loss/wilting, which limits how much further the rate of transpiration can increase, oe
(d) Answer: At higher temperatures, the plant may start to close its stomata to prevent excessive water loss, which limits how much further the rate of water uptake can rise.

Question 12

(a) B1 cutting underwater prevents air from entering the xylem vessels (avoiding an air lock/embolism), which would otherwise block the flow of water and stop accurate water uptake being measured, oe
(a) B1 cutting at a slant increases the surface area of the cut end for water uptake and/or makes the shoot easier to insert into the tubing, oe
(a) Answer: Cutting underwater stops air entering and blocking the xylem vessels, which would prevent accurate measurement of water uptake; cutting at a slant increases the surface area of the cut end and makes the shoot easier to insert.
(b) B1 to make the apparatus airtight/watertight, so that all the water taken up is drawn through the capillary tube (and none is lost or air enters at the joints), giving an accurate measurement of water uptake, oe
(b) Answer: It makes the apparatus airtight and watertight, so all the water taken up passes through the capillary tube, giving an accurate reading of water uptake.
(c) B1 independent variable: humidity (e.g. changed by placing a plastic bag/humidity chamber around the shoot), oe
(c) B1 dependent variable: rate of water uptake (the distance the bubble moves per unit time), oe
(c) B1 controlled variable: any one of temperature, light intensity, air movement, the same species/size of plant or leaf area, oe
(c) Answer: Independent variable: humidity. Dependent variable: rate of water uptake (distance the bubble moves per unit time). Controlled variable: e.g. temperature.
(d) B1 not all of the water taken up by the shoot is lost through transpiration, oe
(d) B1 some of the water is used for other processes in the plant, such as photosynthesis or keeping cells turgid/supported, oe
(d) Answer: A potometer measures water uptake, but not all of this water is lost by transpiration; some is used for other processes, such as photosynthesis and keeping cells turgid, so uptake only approximates transpiration.

Question 13

(a) M1 radius = 1.0 mm; area = π x 1.0², oe
(a) A1 3.14 mm² (accept 3.14-3.15) awrt 3 sf
(a) Answer: 3.14 mm²
(b) M1 volume = (their) area x distance = 3.14 x 45, oe, ft their (a)
(b) dM1 rate = volume / time = their volume / 3.0, oe, dependent on the M1 above
(b) A1 47.1 mm³/min (accept 47.0-47.2) awrt 3 sf, ft
(b) Answer: 47.1 mm³/min
(c) M1 their rate x 60 (to convert per minute to per hour), then divide by 1000 (to convert mm³ to cm³), in either order, ft their (b)
(c) A1 2.83 cm³/hour (accept 2.82-2.84) awrt 3 sf, ft
(c) Answer: 2.83 cm³/hour
(d) M1 new rate = their area x 63 / 3.0 (= 65.9-66.0 mm³/min), oe, ft their (a)
(d) dM1 percentage increase = (new rate - their (b)) / their (b) x 100, oe, dependent on the M1 above, ft
(d) A1 40.0% (accept 39-41%) awrt, ft
(d) Answer: 40.0%
(e) B1 moving air blows away/removes water vapour molecules from just outside the stomata, oe
(e) B1 this maintains a steeper water vapour concentration gradient between the leaf's air spaces and the air outside, increasing the rate of diffusion of water vapour out of the leaf, oe
(e) Answer: Moving air removes water vapour molecules from around the stomata as soon as they diffuse out, maintaining a steep concentration gradient between the inside and outside of the leaf, which increases the rate of diffusion (and therefore transpiration), as shown by the 40.0% increase in uptake with the fan.

Question 14

(a) B1 phloem
(a) Answer: Phloem
(b) B1 a source is a part of the plant that produces or releases sugar, e.g. the leaves in summer, oe
(b) B1 a sink is a part of the plant that uses or stores sugar, e.g. the roots in summer, oe
(b) Answer: A source is a part of the plant that produces or releases sugar, such as the leaves in summer; a sink is a part that uses or stores sugar, such as the roots in summer.
(c) B1 which parts of the plant act as a source or a sink can change over time (e.g. the roots become the source and the shoots become the sink in spring), oe
(c) B1 being able to transport in either direction allows sucrose to be delivered to wherever it is currently needed, increasing the plant's flexibility to respond to its changing needs, oe
(c) Answer: The roles of source and sink can swap over time, as shown by the example, so being able to transport in either direction lets the plant send sucrose to wherever it is currently needed, rather than being limited to one fixed direction.

Question 15

B1 the lower surface of the leaf is in direct contact with water, oe
B1 stomata on the lower surface would be blocked/covered by water, oe
B1 so gas exchange (and water loss) could not happen efficiently through the lower surface, meaning stomata must be on the upper surface, which is exposed to the air, to allow gas exchange, oe
Answer: The lower surface of a water lily leaf is in direct contact with the water, so stomata there would be blocked by water and could not exchange gases; the upper surface is exposed to the air, so stomata are found there instead, to allow gas exchange to still occur.

Question 16

(a) M1 radius = 0.75 mm; area = π x 0.75², oe
(a) A1 1.77 mm² (accept 1.76-1.77) awrt 3 sf
(a) Answer: 1.77 mm²
(b) M1 density = number counted / their area = 6 / their (a), ft their (a)
(b) A1 3.40 stomata/mm² (accept 3.38-3.41) awrt 3 sf, ft
(b) Answer: 3.40 stomata/mm²
(c) B1 the upper surface receives more direct sunlight and is likely to be warmer than the lower surface, oe
(c) B1 so having fewer stomata there reduces the amount of water lost by transpiration from the hottest, most exposed surface, helping the plant conserve water, oe
(c) Answer: The upper surface is exposed to more direct sunlight and is warmer, so having fewer stomata there reduces water loss by transpiration from the hottest surface, helping the plant conserve water.
(d) B1 a plant adapted to a hot, dry habitat is likely to have a lower stomatal density (fewer stomata per mm²), oe
(d) B1 this reduces the rate of water loss by transpiration, which is important where water is scarce, oe
(d) B1 a plant adapted to a cool, wet habitat can have a higher stomatal density, allowing more gas exchange (and a higher rate of photosynthesis), without a high risk of the plant losing too much water, oe
(d) Answer: A plant from a hot, dry habitat is likely to have a lower stomatal density, reducing water loss by transpiration where water is scarce. A plant from a cool, wet habitat can have a higher stomatal density, allowing more gas exchange for photosynthesis, since there is little risk of losing too much water.

Question 17

Level 3 (5-6): A clear, logically linked explanation that covers both why wind increases the rate of transpiration (referring to the water vapour concentration gradient at the leaf surface) and its link to wilting, and the knock-on effect on stomatal closure and photosynthesis rate, using appropriate scientific vocabulary throughout.
Level 2 (3-4): Some correctly linked reasoning showing how wind affects the rate of transpiration (e.g. referring to diffusion or water vapour being removed) and/or a link to wilting, but the explanation is incomplete, or the link to photosynthesis is missing or undeveloped.
Level 1 (1-2): Basic, largely isolated statements, e.g. that wind increases water loss and/or the plant wilts, with little or no linking of ideas and limited use of scientific terms.
Level 0 (0): No relevant content.
Indicative content:
- Transpiration is the loss of water vapour from the leaves, mainly through the stomata.
- Wind blows away/removes water vapour molecules from just outside the stomata (from the leaf surface).
- This maintains, or increases, the difference in water vapour concentration (the diffusion gradient) between the air spaces inside the leaf and the air outside.
- A steeper concentration gradient increases the rate of diffusion of water vapour out through the stomata, increasing the rate of transpiration.
- Faster transpiration means water can be lost from the plant faster than it is being absorbed and transported by the roots and xylem.
- If water loss exceeds water uptake, cells (such as mesophyll and guard cells) lose turgor, and the plant wilts.
- In response, guard cells may cause the stomata to partially or fully close, to reduce further water loss.
- If stomata close, less carbon dioxide can diffuse into the leaf.
- Since carbon dioxide is a raw material needed for photosynthesis, a reduced supply is likely to reduce the rate of photosynthesis.
- Reduced water availability in the leaf cells may also directly limit photosynthesis, since water is also a raw material for the reaction.

Plant Tissues, Transport and Transpiration - Worksheets, Questions and Revision