Biology • Year 12 • Module 5 • Lesson 3
Reproduction in Plants, Fungi, Bacteria and Protists
Apply pollination, vegetative propagation, budding, spores and binary fission to real organisms, real growth data and a real-world scenario.
1. Sequence the steps — sexual reproduction in a flowering plant
The events below describe the full sequence of sexual reproduction in a flowering plant from pollen production to seed formation. They are shuffled. Write the correct order (1–8) in the box beside each event. 8 marks
| Order | Event |
|---|---|
| The pollen grain germinates on the stigma and grows a pollen tube down the style toward the ovule. | |
| A bee carrying pollen lands on a second flower and pollen grains stick to the stigma. | |
| The male gamete fuses with the female gamete in the ovule, forming a diploid zygote. | |
| Pollen grains are produced inside the anther of the stamen. | |
| The ovule develops into a seed and the ovary develops into a fruit. | |
| The fruit is dispersed by an animal, water or wind, and the seed germinates in a new location. | |
| The bee picks up pollen from the anther of one flower while feeding on nectar. | |
| The pollen tube reaches the ovule and delivers the male gamete to the egg cell. |
2. Interpret data — runner growth in commercial strawberries
A horticulture team grew four cultivars of strawberry (Fragaria × ananassa) under identical light, water and nutrient conditions. They recorded the mean number of runners produced per parent plant over 12 weeks, and the percentage of those runners that successfully rooted to form new daughter plants. 8 marks
| Cultivar | Mean runners per parent (12 wk) | Successful daughter plants (%) | Daughter plants per parent |
|---|---|---|---|
| Albion | 9.2 | 88 | 8.1 |
| Camarosa | 14.6 | 72 | 10.5 |
| Festival | 6.4 | 94 | 6.0 |
| San Andreas | 11.0 | 80 | 8.8 |
Data: stylised horticultural trial, NSW strawberry growers' research note (illustrative).
2.1 Identify which cultivar produces the most runners per parent, and which produces the most daughter plants per parent over 12 weeks. 2 marks
2.2 A grower argues: "More runners is always better — pick Camarosa." Using the data, explain why this is too simple and identify what actually determines reproductive output here. 3 marks
2.3 Each daughter plant produced by a runner is a clone of the parent. Explain one commercial advantage and one long-term biological risk of using this kind of asexual reproduction across a 50-hectare strawberry farm. 3 marks
3. Interpret real data — bacterial binary fission growth curve
The figure below shows the population size of Escherichia coli in a nutrient-rich broth at 37 °C, measured over 8 hours. The y-axis is on a log scale (powers of 10). Data adapted from Monod (1949), Annual Review of Microbiology 3: 371–394 — classical bacterial growth kinetics.
Figure adapted from Monod (1949), Annual Review of Microbiology 3, 371–394 — bacterial growth kinetics in batch culture.
3.1 Describe the trend in cell density between 1 h and 5 h of the experiment, naming the type of growth. 2 marks
3.2 The y-axis is on a log scale. Estimate the cell density at 1 h and at 5 h, and calculate approximately how many doublings of the bacterial population occurred between these times. 2 marks
3.3 Using lesson content, explain why binary fission allows this rapid increase in numbers, and suggest one reason the curve plateaus after about 5 hours even though the bacteria are still dividing. 3 marks
4. Case study — the 1845 Irish potato famine
In the 1840s, Irish farmers grew almost exclusively a single potato variety, the "Lumper", which was reproduced asexually from tubers and was therefore genetically near-identical across the country. In 1845, a strain of the oomycete Phytophthora infestans (potato blight) arrived in Ireland. Within two growing seasons most of the national potato crop failed. Approximately one million people died and another million emigrated. Wild and South American potato populations — which reproduce sexually from seed and contain far greater genetic variation — included individuals with natural resistance to P. infestans.
In 4–6 sentences, explain — using lesson content on vegetative propagation, variation and continuity of species — why the Lumper population was so vulnerable and why the wild seed-grown populations were not. 5 marks
Q1 — Correct order of flowering-plant sexual reproduction
- Pollen grains are produced inside the anther of the stamen.
- The bee picks up pollen from the anther of one flower while feeding on nectar.
- A bee carrying pollen lands on a second flower and pollen grains stick to the stigma.
- The pollen grain germinates on the stigma and grows a pollen tube down the style toward the ovule.
- The pollen tube reaches the ovule and delivers the male gamete to the egg cell.
- The male gamete fuses with the female gamete in the ovule, forming a diploid zygote.
- The ovule develops into a seed and the ovary develops into a fruit.
- The fruit is dispersed by an animal, water or wind, and the seed germinates in a new location.
Marking note: award 1 mark per correctly placed event. Steps 2 and 3 together form pollination; step 6 is fertilisation.
Q2 — Strawberry runner data
2.1 Most runners per parent: Camarosa (14.6 runners). Most daughter plants per parent: Camarosa (10.5 daughter plants) — despite a lower success rate, its total runner number is high enough that it still produces the most daughters.
2.2 The argument is too simple because the number of runners on its own does not measure reproductive output. Festival produces only 6.4 runners but 94% root successfully, while Camarosa produces 14.6 runners but only 72% root. What actually determines reproductive output is runners × success rate = daughter plants per parent. So although Camarosa wins here, the comparison must include both runner number and runner survival.
2.3 Advantage: rapid asexual spread of a successful genotype — every daughter inherits the parent's commercially valuable traits (fruit size, flavour, yield) without the variation that sexual reproduction would introduce. Risk: the whole 50-hectare planting is genetically uniform, so a single pathogen, pest or environmental change that affects the parent genotype will affect every daughter plant — leading to potentially catastrophic crop loss (this is the same vulnerability that affected the Cavendish banana and the Lumper potato).
Q3.1 — Trend description
Between 1 h and 5 h the cell density rises rapidly. On the log scale the increase is roughly linear, which means the population is undergoing exponential growth — bacteria are dividing by binary fission at a near-constant rate. This is the "log phase" of the bacterial growth curve.
Q3.2 — Read-off and doublings
At 1 h ≈ 10³ cells mL⁻¹; at 5 h ≈ 10⁹ cells mL⁻¹. The change is from 10³ to 10⁹, a factor of 10⁶ (one million-fold). Since each binary fission doubles the population, the number of doublings is log₂(10⁶) ≈ ~20 doublings across 4 hours (consistent with a generation time of roughly 12 minutes for E. coli under ideal conditions). Accept any answer in the range 18–22 doublings.
Q3.3 — Why binary fission is so fast, and why the curve plateaus
Binary fission is fast because each cell replicates its DNA and divides directly into two daughter cells without producing gametes, mating or fertilisation, so every cell in the population can reproduce simultaneously. The curve plateaus (stationary phase) because the bacteria progressively exhaust nutrients in the broth and accumulate waste products. The rate of new cells produced by binary fission becomes balanced by the rate of cell death, so net cell density stops increasing even though some cells are still dividing.
Q4 — Case study (sample top-band response)
The Lumper potato was reproduced by vegetative propagation from tubers, so every Lumper plant in Ireland was a genetic clone of every other Lumper plant. This means the population had almost no genetic variation, so when Phytophthora infestans arrived no individuals carried alleles for resistance — if the pathogen could infect one Lumper plant, it could infect every Lumper plant. Wild and South American potato populations reproduce sexually from seed, so cross-fertilisation produces variation through new allele combinations, and some individuals carry resistance alleles which allow the population to survive the disease. The Irish case shows the lesson's central trade-off: asexual reproduction is excellent in stable conditions and lets a useful genotype spread fast, but it leaves the species highly vulnerable when conditions change — in this case, when a new pathogen arrives. Sexual reproduction is slower but its variation gives some offspring a much better chance of surviving the change, supporting continuity of species in the long term.
Marking note: 1 mark for naming tuber reproduction as vegetative propagation, 1 mark for identifying low genetic variation in clones, 1 mark for explaining why low variation means uniform susceptibility, 1 mark for contrasting with sexual reproduction generating variation through gamete fusion, 1 mark for explicit link to continuity of species under changing conditions.