CB6 · Plant structures and their functionsTopic 6 — Plant structures and their functions
Photosynthesis, plant transport and water loss.
Revise the key ideas
Photosynthesis and biomass
Plants and algae use photosynthesis to produce food and therefore biomass. They are producers, forming the starting point of many food chains.
Photosynthesis transfers light energy into chemical energy stored in glucose. It is an endothermic process because it takes in energy.
The word equation is carbon dioxide + water → glucose + oxygen. Light energy and chlorophyll are needed, but are not substances consumed in the equation.
The balanced symbol equation is 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂. The carbon in glucose comes from carbon dioxide, rather than from soil minerals.
Photosynthesis takes place in chloroplasts. Chlorophyll is a pigment that absorbs light energy; it is not the name of the organelle.
Plants respire as well as photosynthesise. Respiration occurs in both light and dark, while photosynthesis requires light.
How plants use glucose
Glucose can be used in respiration to transfer energy for cell processes, including active transport and growth.
Plants join glucose molecules together to make starch for storage. Starch is insoluble, so it does not build up as dissolved sugar in the cell and draw in extra water by osmosis.
Plants convert sugars into sucrose for transport to other parts of the plant. Stored starch can be broken down and the sugars used or converted for transport.
Sugars transported to storage organs, such as potato tubers, can be converted into starch and used later.
Glucose can supply material for cellulose in cell walls and for lipid production. Plants also combine carbon-containing substances with nitrogen from nitrate ions to make amino acids and proteins.
Glucose, starch and sucrose have different roles: glucose is useful in respiration, starch is a storage carbohydrate, and sucrose is a transported sugar.
Leaf cells, stomata and guard cells
Palisade cells near the upper surface of many leaves contain numerous chloroplasts. Their position and chloroplasts help them absorb light for photosynthesis.
A stoma is a pore in the leaf surface; stomata is the plural. In many land plants, numerous stomata occur on the lower leaf surface.
A pair of guard cells controls each stoma. When guard cells take up water by osmosis and become turgid, their shape changes and the pore opens.Guard-cell water content changes the pore opening; schematic surface views.
When guard cells lose water and become less turgid, the pore closes. Their opening depends on environmental signals and water availability, rather than simply on water flow starting or stopping in the stem.
During net photosynthesis, carbon dioxide diffuses into the leaf and oxygen diffuses out through stomata. These gases move down their concentration gradients.
Water vapour can also diffuse out through an open stoma. Opening pores allows gas exchange but can increase water loss.
Stomata commonly open in the light and close in darkness, but this is not an absolute rule for every plant or condition. Water transport does not simply stop at night.
Limiting factors and the inverse-square law
A limiting factor is the factor restricting the rate under the current conditions. The main factors considered here are light intensity, carbon dioxide concentration and temperature.
Increasing light intensity increases the rate of photosynthesis while light is the limiting factor. When the graph levels off (forms a plateau), another factor, such as carbon dioxide or temperature, is limiting the rate.Increasing light raises the rate only while light is limiting.
Increasing carbon dioxide concentration can increase the rate when carbon dioxide is limiting. Increasing a factor already in sufficient supply may have little effect.
Photosynthesis involves enzyme-controlled reactions. Increasing temperature initially speeds reactions, but above an optimum the rate may fall as enzymes lose activity or denature.
The limiting factors work together. For example, adding carbon dioxide may let the rate rise above a previous plateau, until another factor becomes limiting. Use the evidence in the question to identify the limiting factor.
For a suitable small light source, light intensity is proportional to 1 ÷ distance². If the distance doubles, intensity becomes one quarter; if it triples, intensity becomes one ninth.
Use I₂ ÷ I₁ = (d₁ ÷ d₂)². Moving a lamp from 10 cm to 20 cm gives I₂ ÷ I₁ = (10 ÷ 20)² = 0.25.
Photosynthesis rate is approximately proportional to light intensity only while light is limiting. Do not assume that halving distance always quadruples the rate at an existing plateau.
Core practical: light intensity and photosynthesis
Place a suitable piece of pondweed in water or a dilute sodium hydrogencarbonate solution, which supplies carbon dioxide. Use a lamp at measured distances to vary light intensity.Vary lamp distance while controlling other conditions; measure oxygen production over time.
Estimate photosynthesis rate by counting bubbles released in a fixed time, or collect the gas and measure its volume per unit time. Oxygen production provides an indicator of photosynthesis rate.
Allow the plant to adjust at each distance before measuring. Repeat readings and calculate a mean to reduce the effect of random variation.
Keep temperature, carbon dioxide supply, pondweed species and size, and measurement time controlled. A lamp can also heat the water; monitor temperature and use an appropriate heat-control arrangement.
Bubble counts are an estimate because bubbles can differ in size. Measuring gas volume is usually a better comparison than assuming that every bubble contains the same amount of gas.
Rate = gas volume ÷ time. For example, 3 cm³ of oxygen collected in 2 minutes gives a rate of 1.5 cm³ per minute.
Plot rate against light intensity, or use 1 ÷ distance² as an estimate of relative intensity when appropriate. Interpret any plateau using limiting factors.
Follow school practical instructions for handling glass, cutting pondweed and using electrical equipment near water. Keep the apparatus and procedure consistent between readings.
Root hair cells and mineral uptake
A root hair cell has a long projection that provides a large surface area for absorbing water and mineral ions from the soil.
A thin cell wall gives a short movement distance. A large surface area relative to volume helps exchange, but the cell membrane still controls what enters the cell.
Water enters across the partially permeable cell membrane by osmosis when the surrounding solution has a higher water concentration than the cell contents.
Mineral ions can be absorbed by active transport when their concentration is lower in the soil than inside the cell. Carrier proteins use energy transferred by respiration.
Nitrate ions supply nitrogen for making amino acids and proteins. Water and mineral ions pass through root tissues into the xylem for transport through the plant.
Water can move through cell-wall spaces as well as across cell membranes on its way through root tissues. Do not describe every step as osmosis; osmosis specifically involves a partially permeable membrane.
Xylem and the transpiration stream
Xylem carries water and mineral ions mainly upwards from roots to stems and leaves. Mature xylem vessel elements are dead and have no cytoplasm obstructing the passage.
The vessel elements join end to end with their end walls lost or broken down, making a continuous hollow tube.
Lignin strengthens xylem vessel walls and supports the plant. It also stops the vessels collapsing as water is pulled upwards under tension.
Water evaporates from moist cell surfaces inside the leaf and water vapour diffuses through stomata into the surrounding air. This loss of water vapour is transpiration.
Evaporation from leaves creates a pull on the continuous water column in the xylem. Water is drawn upwards to replace what is lost, creating the transpiration stream.
Water arriving at leaves supplies photosynthesis, carries mineral ions and helps keep cells turgid. Evaporation can also contribute to cooling.
If a plant loses water faster than it takes it up, its cells become less firm (lose turgor) and it may wilt. Closing stomata reduces water loss but also lets less carbon dioxide enter.
Environmental effects on water loss
Higher temperature usually increases evaporation and the rate of water loss, provided the plant has enough water and stomata remain open.
Air movement removes humid air near the leaf, maintaining a steep water-vapour concentration gradient. Wind can therefore increase transpiration.
Low humidity gives a steeper gradient between moist air inside the leaf and drier external air. High humidity generally reduces water loss.
Light often promotes stomatal opening, increasing gas exchange and water loss. Drought responses or other limiting conditions can change this pattern.
Compare conditions while controlling other variables. Explain each effect using evaporation, stomatal opening or the concentration gradient, rather than just stating that a plant drinks more.
Phloem and translocation
Phloem transports sucrose from sources (where sugar is made or released) to sinks (where sugar is used or stored). This movement of sugar is called translocation.
A leaf carrying out photosynthesis can be a source. Growing roots, fruits and storage organs can be sinks. A storage organ becomes a source when its stored carbohydrate is broken down and released as sugar.
Phloem contains living sieve-tube elements joined together. Sieve plates have pores allowing sap to pass between elements; companion cells support their activity.
Companion cells contain mitochondria and use energy in sugar loading and unloading. Phloem transport depends on living cells, unlike the water-conducting elements of xylem.
Different phloem tubes can carry sugars upwards or downwards according to source and sink locations. Do not treat phloem as an exclusively downward flow.
Compare tissues clearly: xylem transports water and mineral ions through lignified dead vessels; phloem transports sucrose using living tissue and energy.Phloem direction varies between tubes and source–sink arrangements; arrows compare possible directions, not simultaneous opposite flow in one tube.
Measuring water uptake and calculating rates
A potometer measures water uptake by a leafy shoot. Uptake can estimate transpiration because much of the water is lost through leaves, but some is used or retained by the plant.
In a bubble potometer, measure how far an air bubble moves along a capillary tube in a known time. Keep the apparatus airtight and watertight and allow conditions to stabilise.
Movement rate = distance ÷ time. A bubble moving 30 mm in 5 minutes has a rate of 6 mm per minute.
If the capillary cross-sectional area is known, water volume taken up = area × distance moved. Use consistent units before dividing the volume by time.
For a capillary area of 0.5 mm² and a bubble movement of 20 mm, uptake is 10 mm³. Over 2 minutes, the volume uptake rate is 5 mm³ per minute.
Repeat readings under each condition and calculate a mean. Reset the bubble consistently and compare one environmental variable at a time; use results as uptake estimates, not exact direct measurements of evaporation.
Watch CB6 · Plant structures and their functions · Topic 6 — Plant structures and their functions
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