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. All populations are diploid and all data sets are fictional. Allele frequency = copies of the allele ÷ (2 × number of individuals). Hardy–Weinberg equations are not needed.
Question 1 (multiple choice · foundation)
A pond holds a population of 20 snails with two shell-colour alleles. Which event is an example of genetic drift?
- (A) A cow wades into the pond and, by chance, crushes six snails; the survivors have different allele frequencies from the original 20.
- (B) Birds see and eat more pale snails than dark snails, so the dark allele becomes more common.
- (C) Ducks carry snail eggs in from a neighbouring pond, adding alleles that were not present before.
- (D) An error during DNA replication produces a new shell-colour allele in one snail.
Answer and explanation
Answer: (A). The snails killed were not chosen by shell colour; the change in allele frequencies is due to chance in a small population.
- (B) is natural selection: survival depends on the phenotype.
- (C) is gene flow: alleles move in from another population.
- (D) is mutation: a new allele arises. Mutation is random too, but it is a separate process from drift.
Question 2 (multiple choice · core)
A meadow population of 100 plants has these genotypes for a flower-colour gene: 45 RR, 30 Rr and 25 rr. What is the frequency of allele R?
- (A) 0.30
- (B) 0.45
- (C) 0.60
- (D) 0.75
Answer and explanation
Answer: (C). Total alleles = 2 × 100 = 200. Copies of R = (2 × 45) + 30 = 120. Frequency = 120 ÷ 200 = 0.60.
- (A) is the frequency of heterozygotes (30 ÷ 100), not of an allele.
- (B) is the frequency of RR plants only, ignoring the R alleles in heterozygotes.
- (D) is the fraction of plants carrying at least one R ((45 + 30) ÷ 100). It counts individuals, not alleles.
Question 3 (multiple choice · core)
Two lakes are joined by a stream, and fish of one species swim freely between them. A dam is built that blocks the stream completely. Which prediction is best supported?
- (A) Over many generations, allele frequencies in the two lake populations will become more different.
- (B) Allele frequencies in the two lakes will become identical, because the fish are one species.
- (C) Both populations will stop evolving, because no new alleles can arrive.
- (D) Mutation rates will rise in both lakes to replace the alleles no longer arriving.
Answer and explanation
Answer: (A). Gene flow had kept the two populations similar. Without it, drift, mutation and any selection act on each lake separately, so their allele frequencies drift apart.
- (B) reverses the effect: it was gene flow that kept them alike; removing it allows divergence.
- (C) is wrong because drift, mutation and selection all continue within each lake.
- (D) is wrong because mutation is random and does not respond to need.
Question 4 (investigation · core)
A class sets up 12 small vials, each started with 10 fruit flies, and 12 large cages, each started with 500 flies. In every population, a neutral marker allele (one with no effect on survival or reproduction) starts at a frequency of 0.50. Each generation, offspring are moved to a fresh vial or cage, keeping the population size the same. After 20 generations the frequencies are:
| Population | Marker allele frequencies after 20 generations |
|---|---|
| 12 small vials (10 flies) | 0.00, 0.15, 1.00, 0.35, 0.00, 0.80, 0.55, 1.00, 0.00, 0.25, 0.90, 0.65 |
| 12 large cages (500 flies) | 0.43, 0.58, 0.47, 0.52, 0.61, 0.41, 0.50, 0.55, 0.45, 0.63, 0.49, 0.54 |
(a) State a null hypothesis for the effect of population size on change in allele frequency. (b) Before the results, predict what drift would do in each group. (c) Describe the results, using the range and the number of populations where the allele was fixed or lost. (d) Explain why the class used a neutral marker.
Model answer and suggested Marlbridge rubric
(a) “Population size has no effect on how much allele frequencies change; small and large populations will show the same spread of final frequencies.”
(b) Small populations: frequencies wander widely in random directions; some reach 0 or 1. Large populations: frequencies stay close to 0.50. The average across replicates should stay near 0.50 in both groups, because drift has no preferred direction.
(c) Small vials ranged from 0.00 to 1.00 (range 1.00): the allele was lost in 3 vials and fixed in 2. Large cages ranged only from 0.41 to 0.63 (range 0.22), with none fixed or lost. The means were similar (about 0.47 and 0.52). The results reject the null hypothesis: the spread is far larger in small populations.
(d) With a neutral allele, selection cannot explain any change, so differences can be put down to chance (drift). A marker affecting survival would mix selection with drift.
| Point | What earns it |
|---|---|
| 1 | Null hypothesis of no effect of size on allele-frequency change |
| 1 | Prediction: wide random spread, fixation/loss in small; little change in large |
| 1 | Description with both ranges and counts of fixed (2) and lost (3) |
| 1 | Neutral marker removes selection as an explanation |
Question 5 (calculation · core)
A beetle species on a mainland has a gene for body colour with alleles B (brown) and b (black, recessive). A mainland sample of 100 beetles: 64 BB, 32 Bb, 4 bb. Years ago, a few beetles were blown to an island. Today a sample of 40 island beetles: 10 BB, 14 Bb, 16 bb.
(a) Calculate the frequency of b on the mainland and on the island. (b) Explain how a founder effect could produce this difference. (c) A student says black beetles must survive better on the island. State a null hypothesis and describe evidence that would favour selection over drift.
Worked solution and suggested Marlbridge rubric
(a) Mainland: copies of b = (2 × 4) + 32 = 40, out of 200. Frequency of b = 0.20. Island: copies of b = (2 × 16) + 14 = 46, out of 80. Frequency of b = 0.575 (nearly three times the mainland value).
(b) The island population came from only a few founders. By chance, they carried a higher proportion of b alleles than the mainland population. The island gene pool started with the founders’ frequencies, and drift in the small early population could have pushed b further. No selection is needed to explain the difference.
(c) Null hypothesis: “Survival and reproduction on the island do not depend on body colour.” Evidence for selection: black beetles survive or reproduce better than brown ones on the island in a mark-and-recapture study; b rises consistently over several generations; several separate islands with the same conditions all show high b. Evidence for drift instead: other neutral genes also differ strongly between island and mainland, and different islands vary in random directions.
| Point | What earns it |
|---|---|
| 1 | Both frequencies correct, with working |
| 1 | Founders a small, chance sample, so frequencies differ from the source |
| 1 | Appropriate null hypothesis |
| 1 | One valid line of evidence that would distinguish selection from drift |
Question 6 (data analysis · stretch)
A fictional ground-nesting bird lives only in one reserve. Habitat loss reduced it to about 50 birds. In 2005, managers released 30 birds from a large, distant population of the same species.
| Year | 1970 | 1995 | 2015 |
|---|---|---|---|
| Population size | about 2000 | about 50 | about 160 |
| Eggs that hatch / % | 90 | 52 | 81 |
| Mean number of alleles per gene (15 genes) | 5.1 | 2.4 | 3.8 |
(a) Describe the changes from 1970 to 1995, with percentage changes. (b) Explain how the bottleneck could have lowered hatching success. (c) Explain the changes between 1995 and 2015. (d) Give one limitation of concluding that the release caused the recovery.
Model answer and suggested Marlbridge rubric
(a) The population fell from about 2000 to about 50. Hatching success fell from 90% to 52% ((52 − 90) ÷ 90 × 100 = −42.2%). Mean alleles per gene fell from 5.1 to 2.4 (−52.9%).
(b) In the small population, drift removed many alleles and the surviving birds were closely related. Harmful recessive alleles, once rare, could become common by chance, and mating between relatives produced more embryos homozygous for them, so more eggs failed to hatch.
(c) The released birds brought new alleles (gene flow), raising mean alleles per gene from 2.4 to 3.8 (+58.3%). Fewer offspring were homozygous for harmful recessive alleles, so hatching rose from 52% to 81% (+55.8%), and the population grew to about 160. This is genetic rescue.
(d) Any one: there is no comparison population without a release; habitat management or predator control may also have changed after 1995; only three time points are given; hatching may vary between years for other reasons (weather).
| Point | What earns it |
|---|---|
| 1 | Changes described with −42.2% (hatching) and −52.9% (alleles) |
| 1 | Bottleneck → loss of variation and more homozygous harmful recessives |
| 1 | Gene flow restores alleles, linked to improved hatching |
| 1 | A valid limitation |
Question 7 (calculation · stretch)
For one gene in a population of 40 000 diploid animals, the chance that any given allele copy mutates to a new form in one generation is 2 × 10⁻⁵.
(a) Calculate the expected number of new mutant alleles of this gene per generation. (b) Calculate the frequency of one new mutant allele when it first appears in this population, and in a population of 20 animals. (c) Use your answers to explain why mutation is the source of variation but rarely, by itself, changes allele frequencies much, and why the fate of a new allele depends on population size.
Worked solution
(a) Allele copies = 2 × 40 000 = 80 000. Expected new mutants = 80 000 × 2 × 10⁻⁵ = 1.6 per generation.
(b) One copy among 80 000: frequency = 1 ÷ 80 000 = 1.25 × 10⁻⁵. In 20 animals: 1 ÷ 40 = 0.025.
(c) Mutation does create new alleles every generation (about 1.6 here), so it supplies the variation on which selection and drift act. But each new allele starts at a tiny frequency, so mutation alone moves frequencies very little. What happens next depends on chance and selection: in a large population a new neutral allele stays rare for a long time, while in a small population it already starts at 0.025 and drift can make it common, or lose it, within a few generations.
Suggested mark points (3): 1 for 1.6; 1 for both frequencies; 1 for linking low starting frequency to slow change by mutation alone and stronger drift in small populations.
Common error: using 40 000 instead of 80 000 copies gives 0.8 new mutants and a starting frequency of 2.5 × 10⁻⁵.
How did you do?
- Q1 or Q3 wrong: re-read “Genetic drift” and “Gene flow” in the study guide, and the summary table.
- Q2, Q5 or Q7 calculations wrong: rework Worked examples 1 and 2; count alleles, doubling homozygotes.
- Q4 incomplete: re-read “Reading Figure 1” and the null hypothesis in Worked example 1.
- Q6 incomplete: re-read “The bottleneck effect” and the paragraph on genetic rescue.
Then tick off the topic checklist.