These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Useful facts: a gamete carries half the chromosomes of the plant that made it; an odd number of chromosome sets, or unmatched chromosomes, usually makes an organism sterile. All species and data sets are fictional.
Question 1 (multiple choice · foundation)
Two fictional tree frog species live around the same ponds. Species P breeds only in March and species Q only in May. In a laboratory, their eggs and sperm can be mixed to make healthy, fertile tadpoles. Which barrier keeps the two species apart in nature?
- (A) Temporal isolation, a pre-zygotic barrier
- (B) Habitat isolation, a pre-zygotic barrier
- (C) Mechanical isolation, a pre-zygotic barrier
- (D) Reduced hybrid viability, a post-zygotic barrier
Answer and explanation
Answer: (A). The frogs share ponds but breed at different times of year, so their gametes never meet in nature. That is temporal isolation, and it acts before a zygote can form.
- (B) Habitat isolation needs different habitats. These frogs use the same ponds.
- (C) Mechanical isolation means reproductive parts do not fit. Nothing suggests this, and frogs fertilize eggs outside the body.
- (D) The hybrids are healthy and fertile in the laboratory, so there is no post-zygotic barrier.
Question 2 (multiple choice · core)
Which scenario is most likely to lead to sympatric speciation?
- (A) A glacier advances across a valley and splits a population of voles into two.
- (B) A storm carries a few birds to an island 300 km from the mainland.
- (C) An error in cell division produces a tetraploid grass plant in the middle of a diploid meadow.
- (D) A river changes course and divides a forest beetle population.
Answer and explanation
Answer: (C). The tetraploid grows among its diploid parents, so there is no geographic separation. Crosses with diploids give sterile triploids, so the tetraploid line is isolated at once.
- (A) A glacier is a geographic barrier: allopatric.
- (B) An island far from the mainland is geographically isolated: allopatric (a founder population).
- (D) A river dividing a population is a geographic barrier: allopatric.
Question 3 (multiple choice · core)
Fossil shells of a fictional freshwater snail show almost no change in shape through rock layers spanning 2 million years. Then, within layers representing about 20,000 years, a new, wider shell form appears and persists unchanged for the next 1.5 million years. Which model best fits these data?
- (A) Gradualism, because the change took 20,000 years rather than one generation
- (B) Punctuated equilibrium, because long stasis is interrupted by a short period of rapid change
- (C) Convergent evolution, because the new form looks like other wide-shelled snails
- (D) Gradualism, because the total record covers millions of years
Answer and explanation
Answer: (B). The pattern is long stasis (2 million and then 1.5 million years with little change) separated by a change lasting about 1% of the earlier stable period (20,000 ÷ 2,000,000). That is punctuated equilibrium.
- (A) Punctuated equilibrium does not claim change happens in one generation. A few thousand years is “rapid” in geological time.
- (C) Convergent evolution compares unrelated lineages. These data follow one lineage through time.
- (D) A long record does not make change gradual. What matters is whether change was steady or concentrated in short bursts.
Question 4 (calculation · core)
Two fictional wild plant species grow together on a hillside. Species K has 2n = 10; species L has 2n = 14. Insects sometimes carry pollen of K to flowers of L.
(a) Give the chromosome number of a K × L hybrid and explain why it is sterile. (b) In one hybrid, the chromosome number doubles. Give the new chromosome number and explain why this plant is fertile. (c) Give the chromosome number of the offspring if the doubled plant is crossed with species K, and say whether that offspring is likely to be fertile. (d) Explain whether the doubled plant is a new species and what type of speciation this is.
Worked solution and suggested Marlbridge rubric
(a) K gamete 10 ÷ 2 = 5; L gamete 14 ÷ 2 = 7. Hybrid: 5 + 7 = 12. Its K chromosomes and L chromosomes are different, so in meiosis most chromosomes have no matching partner to pair with. Gametes get unbalanced sets and fail: the hybrid is sterile (reduced hybrid fertility).
(b) 12 × 2 = 24. Every chromosome now has an identical partner, so pairing in meiosis works and gametes carry 12 chromosomes (one full K set plus one full L set). The plant is fertile.
(c) 12 + 5 = 17. The L chromosomes have no partners, so the offspring is likely to be sterile.
(d) The 24-chromosome plants breed with each other but produce sterile offspring with K (and with L: 12 + 7 = 19). They are reproductively isolated, so they are a new species under the biological species concept. Because it arose among its parents on the same hillside, it is sympatric speciation (by polyploidy).
| Point | What earns it |
|---|---|
| 1 | 12 chromosomes and sterility explained by unpaired chromosomes in meiosis |
| 1 | 24 chromosomes and fertility explained by every chromosome having a partner |
| 1 | 17 chromosomes, likely sterile |
| 1 | New species justified by reproductive isolation from both parents |
| 1 | Sympatric, with the reason (same area, no geographic barrier) |
Common error: adding the parents’ diploid numbers (10 + 14 = 24) gives the right number in (b) by coincidence, but it skips the hybrid step; it earns the mark only with the reasoning.
Question 5 (data analysis · core)
Two fictional species of reef worm, X and Y, release eggs and sperm into the sea on the same nights. In a laboratory, eggs of each species were mixed with sperm of each species, 500 eggs per trial.
| Eggs | Sperm | Eggs fertilized |
|---|---|---|
| X | X | 470 |
| X | Y | 15 |
| Y | Y | 460 |
| Y | X | 10 |
(a) Calculate the percentage fertilized in each trial. (b) Identify the barrier shown and classify it as pre-zygotic or post-zygotic. (c) Explain why this barrier is especially important for animals that spawn into open water. (d) A mutation changes the egg-surface protein of some X worms so that their eggs bind Y sperm as well as X sperm. Predict the effect on gene flow between X and Y.
Model answer and suggested Marlbridge rubric
(a) X × X: 470 ÷ 500 × 100 = 94.0%. X eggs + Y sperm: 3.0%. Y × Y: 92.0%. Y eggs + X sperm: 2.0%.
(b) Sperm rarely fertilize eggs of the other species, even when gametes are mixed: gametic isolation. It is pre-zygotic, because it stops a zygote forming (it acts after the gametes meet, but before fertilization).
(c) These worms spawn on the same nights in the same water, so there is no habitat, timing, courtship or mechanical barrier. Eggs and sperm of both species mix freely. Matching between egg and sperm is the first point where the species can be kept apart.
(d) More X eggs would be fertilized by Y sperm, so more hybrid zygotes would form. Gene flow increases only if the hybrids survive and are fertile. If they are, alleles pass between the species and the gene pools could start to merge; if the hybrids die or are sterile, a post-zygotic barrier still blocks gene flow (but X worms waste eggs).
| Point | What earns it |
|---|---|
| 1 | All four percentages correct |
| 1 | Gametic isolation, pre-zygotic, with reason |
| 1 | Explains that other barriers are absent because gametes mix in open water |
| 1 | Predicts more hybrid zygotes |
| 1 | Links gene flow to hybrid viability and fertility |
Question 6 (data analysis · core)
A fictional genus of flightless weevils colonised a chain of volcanic islands. The table shows each island’s age and the number of weevil species found only on that island.
| Island | A | B | C | D |
|---|---|---|---|---|
| Age / million years | 0.4 | 1.1 | 2.3 | 3.8 |
| Weevil species | 3 | 7 | 12 | 13 |
(a) Describe the relationship between island age and species number, using numbers. (b) Calculate the species per million years for islands A and D, and suggest why they differ. (c) Explain why speciation in this genus is an example of adaptive radiation. (d) On two islands, different weevil lineages that are not closely related have both evolved long snouts for feeding inside deep flowers. Name this pattern and explain how it arises.
Model answer and suggested Marlbridge rubric
(a) Older islands have more species: 3 on the 0.4-million-year-old island, rising to 12 on island C. The increase then levels off: island D is 1.5 million years older than C but has only one more species (13).
(b) A: 3 ÷ 0.4 = 7.5 species per million years. D: 13 ÷ 3.8 = 3.4 species per million years. On a young island, many habitats are empty, so new species form quickly. On an old island, most niches are already filled and competition is high, so fewer new species can establish. (This assumes each island was colonised soon after it formed.)
(c) One ancestral lineage gave rise to many species in a short time, as new habitats (new islands, and different habitats on each) became available. Populations adapting to different habitats diverge: this is divergent evolution at a high rate.
(d) Convergent evolution. Feeding in deep flowers is a similar selection pressure on both islands. Individuals with longer snouts reach more food and leave more offspring, so long snouts evolved separately in each lineage. The trait is analogous, not inherited from a shared long-snouted ancestor.
| Point | What earns it |
|---|---|
| 1 | Positive relationship with figures and the levelling off |
| 1 | 7.5 and 3.4 species per million years |
| 1 | Difference explained by empty niches on young islands versus filled niches on old ones |
| 1 | Adaptive radiation: many species from one ancestor as new habitats open |
| 1 | Convergent evolution explained by a similar selection pressure in unrelated lineages |
Question 7 (prediction · stretch)
Two fictional fish species live in the same lake. Males of species B are blue; males of species R are red. Females choose mates by colour. In a laboratory, B × R hybrids are healthy and fertile. Researchers tested mate choice in clear light and in murky (cloudy) water, 60 females per test:
| Water | Females that chose a male of their own species |
|---|---|
| Clear | 58 |
| Murky | 33 |
(a) Identify the barrier that keeps B and R apart in the clear lake. (b) Calculate the percentage of same-species choices in each condition and explain the difference. (c) Run-off from farms is making the lake permanently murky. Predict the effect on gene flow and on the two species over many generations. (d) A student concludes that B and R “were never really separate species”. Evaluate this conclusion using the biological species concept.
Model answer and suggested Marlbridge rubric
(a) Behavioural isolation (a pre-zygotic barrier): females recognise and choose males of their own colour.
(b) Clear: 58 ÷ 60 × 100 = 96.7%. Murky: 33 ÷ 60 × 100 = 55.0%, close to the 50% expected if females chose at random. In murky water females cannot see male colour, so they cannot use it to choose.
(c) More females would mate with males of the other species. Because hybrids are viable and fertile, alleles would flow between the gene pools. Over generations, B and R could merge into a single, more variable hybrid population, and the two species would be lost.
(d) The conclusion is not justified. In the clear lake, mate choice kept B and R almost completely apart (96.7% of choices were for the female’s own species), so very little gene flow happened in nature. That meets the biological species concept, which is about interbreeding in nature. The laboratory hybrids show that their isolation depended on a single pre-zygotic barrier, which makes them vulnerable, but not that they were never separate. The murky-water result does show that isolation can break down, so speciation can be reversed.
| Point | What earns it |
|---|---|
| 1 | Behavioural, pre-zygotic |
| 1 | 96.7% and 55.0%, with the link to seeing colour |
| 1 | Predicts more interspecies matings and increased gene flow |
| 1 | Predicts merging of the gene pools, using the fact that hybrids are fertile |
| 1 | Evaluates the claim using “interbreeding in nature” |
How did you do?
- Q1 or Q5 wrong: re-read the barriers table in the study guide; check when each barrier acts relative to the zygote.
- Q2 or Q4 wrong: re-read “Allopatric and sympatric speciation” and rework Worked example 1.
- Q3 or Q6 wrong: re-read “How fast does speciation happen?” and Figure 2.
- Q7 incomplete: rework Worked example 2 (d), which also predicts what happens when a barrier disappears.
Then tick off the topic checklist.