Pearson Edexcel GCSE Combined Science · B6

Plant structures and their functions

Revision questions with answers for B6 Plant structures and their functions in Pearson Edexcel GCSE Combined Science (spec 1SC0). Practise with flashcards, a topic quiz, a random quiz across the course, 5 crosswords, or read every question and answer below.

All questions and answers

Question 1 · Spec B6.1
Why are green plants and algae described as producers?
Answer:
They make organic food molecules and biomass through photosynthesis.
Question 2 · Spec B6.1
What is biomass?
Answer:
The mass of living or recently living biological material; photosynthesis supplies much of the material in food chains.
Question 3 · Spec B6.2
State the word equation for photosynthesis.
Answer:
Carbon dioxide + water → glucose + oxygen, using light energy.
Question 4 · Spec B6.2
Why is photosynthesis endothermic?
Answer:
It takes in energy from the surroundings, transferred by light.
Question 5 · Spec B6.2
Where does photosynthesis occur in a plant cell?
Answer:
In chloroplasts containing chlorophyll, which absorbs light.
Question 6 · Spec B6.2
Why does photosynthesis increase plant biomass?
Answer:
It converts carbon dioxide and water into organic molecules that supply material for growth.
Question 7 · Spec B6.3
Name three environmental limiting factors for photosynthesis.
Answer:
Temperature, light intensity and carbon dioxide concentration.
Question 8 · Spec B6.3
What is a limiting factor?
Answer:
A factor in shortest effective supply that restricts the rate of a process.
Question 9 · Spec B6.3
Why does increasing light raise photosynthesis rate when light is limiting?
Answer:
More light energy is available to drive the reaction.
Question 10 · Spec B6.3
Why does increasing carbon dioxide sometimes increase photosynthesis?
Answer:
It supplies more of a reactant that may be limiting the rate.
Question 11 · Spec B6.3
Why does photosynthesis slow at very high temperatures?
Answer:
Enzymes involved lose their functional active-site shapes through denaturation.
Question 12 · Spec B6.3
What does a plateau in a light-intensity rate graph show?
Answer:
Further light increases do not increase the rate under those conditions.
Question 13 · Spec B6.4 · Higher only
How can two curves at different carbon dioxide levels reveal limitation?
Answer:
A higher rate at increased carbon dioxide, with other conditions equal, shows carbon dioxide limited the lower curve.
Question 14 · Spec B6.4 · Higher only
Why might heating have little effect when light is very low?
Answer:
Light may limit the rate, so increasing temperature alone cannot substantially increase photosynthesis.
Question 15 · Spec B6.4 · Higher only
How could a greenhouse grower choose which factor to improve?
Answer:
Compare rates under controlled changes to identify the limiting factor, then weigh the extra crop value against costs.
Question 16 · Spec B6.5 · Core practical
Outline a pondweed investigation of light intensity.
Answer:
Change lamp distance and measure oxygen output per unit time while controlling temperature and carbon dioxide availability.
Question 17 · Spec B6.5 · Core practical
Why measure oxygen volume rather than only counting bubbles?
Answer:
Bubble sizes vary, so volume better represents the amount of oxygen produced.
Question 18 · Spec B6.5 · Core practical
Why allow pondweed to adjust after changing lamp distance?
Answer:
Its rate needs time to stabilise before measurement.
Question 19 · Spec B6.5 · Core practical
How can a lamp unintentionally affect another variable?
Answer:
It can warm the water, so temperature must be controlled or monitored.
Question 20 · Spec B6.5 · Core practical
Why keep the pondweed species and amount consistent?
Answer:
Different species or quantities may produce different oxygen outputs independently of light intensity.
Question 21 · Spec B6.5 · Core practical
Pondweed produces 8.0 cm³ gas in 5.0 min. Find rate.
Answer:
8.0 / 5.0 = 1.6 cm³/min.
Question 22 · Spec B6.5 · Core practical
Why repeat measurements at each lamp distance?
Answer:
To assess variation, investigate anomalies and calculate a more reliable mean.
Question 23 · Spec B6.6 · Higher only
State the inverse-square relationship for light intensity.
Answer:
Intensity is proportional to 1 / distance² for a suitable point-source approximation.
Question 24 · Spec B6.6 · Higher only
What happens to intensity if distance doubles?
Answer:
It becomes one-quarter of the original value.
Question 25 · Spec B6.6 · Higher only
What happens to intensity if distance is halved?
Answer:
It becomes four times greater.
Question 26 · Spec B6.6 · Higher only
A lamp moves from 10 to 30 cm away. Find the relative intensity.
Answer:
Distance triples, so intensity becomes 1/9 of the original.
Question 27 · Spec B6.7
How is a root hair cell adapted for absorption?
Answer:
A long projection gives a large surface area and its thin wall provides a short uptake pathway.
Question 28 · Spec B6.7
How does water enter root hair cells?
Answer:
By osmosis through their partially permeable membranes.
Question 29 · Spec B6.7
Why may mineral-ion absorption require energy?
Answer:
Ions may move against their concentration gradient by active transport.
Question 30 · Spec B6.8
What does xylem transport?
Answer:
Water and dissolved mineral ions from roots through the plant.
Question 31 · Spec B6.8
How are xylem vessels adapted to transport?
Answer:
Dead hollow cells form continuous tubes; lignified walls strengthen them.
Question 32 · Spec B6.8
Why is lignin useful in xylem walls?
Answer:
It strengthens the vessels and helps prevent collapse as water is pulled through them.
Question 33 · Spec B6.8
What does phloem transport?
Answer:
Sucrose and other dissolved organic substances between parts of the plant.
Question 34 · Spec B6.8
How does phloem differ from xylem?
Answer:
Phloem contains living transport cells and uses energy in transport; mature xylem vessels are dead and carry water and minerals.
Question 35 · Spec B6.10
What is translocation?
Answer:
Transport of sucrose through phloem from sources, such as leaves, to regions of use or storage.
Question 36 · Spec B6.10
Why can sucrose move towards roots or growing shoots?
Answer:
Both may need sugars for respiration, growth or storage; transport follows the plant's changing needs.
Question 37 · Spec B6.9
Define transpiration.
Answer:
Loss of water vapour from a plant, mainly through stomata in leaves.
Question 38 · Spec B6.9
How does transpiration help water transport?
Answer:
Water loss from leaves creates a pull that draws water up xylem from the roots.
Question 39 · Spec B6.9
What are stomata?
Answer:
Leaf pores through which gases and water vapour diffuse.
Question 40 · Spec B6.9
What is the role of guard cells?
Answer:
They alter the opening of stomata, regulating gas exchange and water loss.
Question 41 · Spec B6.9
Why can closing stomata reduce photosynthesis as well as water loss?
Answer:
It restricts carbon dioxide entry while reducing water-vapour escape.
Question 42 · Spec B6.12
Why does higher temperature usually increase water uptake?
Answer:
It increases evaporation and water loss from leaves, increasing replacement water uptake.
Question 43 · Spec B6.12
Why can moving air increase transpiration?
Answer:
It removes humid air near the leaf, maintaining a steep water-vapour gradient.
Question 44 · Spec B6.12
Why does greater light often increase transpiration?
Answer:
Stomata commonly open in light, increasing the pathway for water-vapour loss.
Question 45 · Spec B6.12
What does a potometer actually measure?
Answer:
Water uptake, used as an estimate of transpiration rather than a direct measure of all water loss.
Question 46 · Spec B6.12
Why must a potometer be airtight?
Answer:
Leaks can move the bubble or admit air, making water uptake estimates unreliable.
Question 47 · Spec B6.13
A bubble moves 30 mm in 5 min. Find its movement rate.
Answer:
30 / 5 = 6 mm/min.
Question 48 · Spec B6.13
A plant takes up 0.60 cm³ water in 10 min. Find uptake rate.
Answer:
0.60 / 10 = 0.060 cm³/min.
Question 49 · Spec B6.12
How can wind's effect on uptake be tested fairly?
Answer:
Vary air movement while keeping light, temperature, plant material and measurement duration comparable.
Question 50 · Spec B6.13
Why state units when comparing transpiration rates?
Answer:
Rates may describe bubble distance or water volume per time; values cannot be compared directly unless measures and units match.

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