Reproduction in plants: IGCSE Biology questions with answers
Free Reproduction in plants questions for IGCSE Biology (0610), part of Reproduction. Each question has the answer and the marking points an examiner would look for.
Insect-pollinated and wind-pollinated flowers have different features.
(a)
Describe two features of a wind-pollinated flower.
[2]
(b)
Define pollination.
[2]
(c)
Describe what happens after a pollen grain lands on the stigma.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
small, dull, no nectar; feathery stigmas; large anthers outside; light pollen
Any 2 of: small, dull petals; anthers hanging outside; feathery stigmas; large amounts of light pollen B2
2
(b)
transfer of pollen from anther to stigma
the transfer of pollen grains from the anther B1 to the stigma B1
2
(c)
pollen tube grows, nucleus fuses with ovule
a pollen tube grows down the style B1 into the ovule B1 the male nucleus fuses with the female nucleus (fertilisation) B1
3
Question 3Structured · 11 marks
This question is about reproduction in flowering plants.
(a)
Define pollination.
[1]
(b)
Describe two differences between insect-pollinated and wind-pollinated flowers.
[2]
(c)
Describe what happens after a pollen grain lands on the stigma, leading to fertilisation.
[3]
(d)
Distinguish between self-pollination and cross-pollination.
[2]
(e)
Explain why cross-pollination is an advantage to a species if the environment changes.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
transfer of pollen from anther to stigma
the transfer of pollen grains from an anther to a stigma B1
1
(b)
insect: large coloured petals, nectar; wind: small dull, feathery stigma outside, anthers hang out
Any 2 of: insect-pollinated have large coloured / scented petals and nectaries, wind-pollinated have small dull petals; wind-pollinated have feathery stigmas / anthers hanging outside the flower; wind-pollinated produce large amounts of light pollen B2
2
(c)
pollen tube grows down the style; to the ovule through micropyle; male nucleus fuses with female nucleus
a pollen tube grows from the pollen grain down through the style B1 it enters the ovule (through the micropyle) B1 the male nucleus fuses with the female nucleus (fertilisation) B1
3
(d)
self: same flower/plant; cross: different plant of same species
self-pollination: pollen from the same flower or plant B1 cross-pollination: pollen from a different plant of the same species B1
2
(e)
more genetic variation; some offspring may have advantageous features; species more likely to survive
cross-pollination produces more genetic variation B1 some offspring are more likely to have features suited to the new conditions B1 so the species is more likely to survive (adapt) B1
3
Question 4Structured · 4 marks
Seeds need certain conditions to germinate.
(a)
State three conditions needed for germination.
[3]
(b)
Explain why oxygen is needed.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
water, oxygen, warmth
water B1 oxygen B1 suitable temperature B1
3
(b)
aerobic respiration for energy
for aerobic respiration to release energy for growth B1
1
Question 5Structured · 5 marks
An insect-pollinated flower has large, brightly coloured petals.
(a)
Explain how the petals help pollination.
[2]
(b)
Distinguish between self-pollination and cross-pollination.
[2]
(c)
Give one disadvantage of self-pollination.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
attract insects
bright colours/scent attract insects B1 which carry pollen to other flowers B1
2
(b)
same plant vs different plant
self: pollen transferred to stigma of same flower/plant B1 cross: to a flower on a different plant of the same species B1
2
(c)
less variation
less genetic variation B1
1
Question 6Structured · 11 marks
A student investigates the conditions needed for germination. Each dish has 20 cress seeds. The number of seeds that germinate after 5 days is recorded in Table 6.1.
dish
repeat 1
repeat 2
repeat 3
A: water, 20 °C, air
18
17
18
B: water, 20 °C, no oxygen
3
2
3
C: water, 5 °C, air
12
13
9
Table 6.1
(a)
Calculate the mean number of seeds that germinated in dish C.
[1]
(b)
Calculate the percentage of seeds that germinated in dish A.
[2]
(c)
Explain the results for dish B.
[2]
(d)
Explain why fewer seeds germinated in dish C than in dish A.
[2]
(e)
Identify the dish that acts as a control and explain why it is needed.
[2]
(f)
Suggest why the student used 20 seeds in each dish and repeated each dish three times.
[1]
(g)
Explain why water was added to every dish.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
11.3
(12 + 13 + 9) ÷ 3 = 11.3 B1
1
(b)
88.3 %
mean = 17.7 C1 17.7 ÷ 20 × 100 = 88.3 % A1
2
(c)
no oxygen; no aerobic respiration; no energy for growth
without oxygen the seeds cannot respire aerobically B1 so there is not enough energy for cell division / growth (allow: a few germinated because some oxygen remained / anaerobic respiration releases a little energy) B1
2
(d)
low temperature; enzymes slow; less respiration
at 5 °C enzymes (e.g. amylase breaking down starch stores) work slowly B1 so less glucose and less respiration / energy for growth B1
2
(e)
A; shows germination when all conditions present
dish A B1 it shows that seeds germinate when all conditions are present, so the effect of removing one condition can be compared B1
2
(f)
reliability, seeds vary
individual seeds vary (some are not viable), so many seeds and repeats give a more reliable result B1
1
(g)
water needed for germination
Any 1 of: water is needed to activate enzymes / for the seed to swell / to transport dissolved food; allow: so that water is not a limiting factor / only oxygen or temperature differs (control variable) B1
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in