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. No constants are needed. Hardy–Weinberg equations: p + q = 1 and p² + 2pq + q² = 1. All organisms’ data sets are fictional.
Question 1 (multiple choice · foundation)
A new virus that infects one species of rabbit reaches four separate populations. Which population is most likely to persist?
- (A) 5000 rabbits, all descended from 10 rabbits released on an island 40 years ago, with no rabbits arriving since
- (B) 200 rabbits in a fenced reserve, most of them the offspring of matings between close relatives
- (C) 3000 rabbits with many different alleles at genes for immune proteins, and regular arrivals from neighbouring populations
- (D) 6000 rabbits that are almost all homozygous for the same immune-gene alleles after strong selection by an earlier virus
Answer and explanation
Answer: (C). Many immune-gene alleles, topped up by gene flow, make it more likely that some rabbits already resist the new virus. They survive, breed and rebuild the population.
- (A) is large, but 10 founders carried only a small sample of the species’ alleles, and nothing has been added since. Size does not equal diversity.
- (B) is small and inbred, so diversity is low.
- (D) is the largest, but selection by the earlier virus left it nearly uniform at these genes; those alleles may not protect against the new virus.
Question 2 (multiple choice · core)
In a coastal population of 500 plants, a recessive allele t gives tolerance to sea-water flooding. Its frequency is 0.4. A storm surge floods the site, and only tt plants survive. Assuming Hardy–Weinberg equilibrium before the storm, about how many plants survive?
- (A) 200
- (B) 80
- (C) 240
- (D) 420
Answer and explanation
Answer: (B). Only the homozygous recessive genotype survives. Its frequency is q² = 0.4² = 0.16. Survivors = 0.16 × 500 = 80.
- (A) uses q instead of q²: 0.4 × 500 = 200. That is the allele frequency, not a genotype frequency.
- (C) uses 2pq = 2 × 0.6 × 0.4 = 0.48, giving 240. Heterozygotes do not survive, because t is recessive.
- (D) uses 1 − q² = 0.84, giving 420. That is the number of plants that die.
Question 3 (multiple choice · core)
In a fictional mouse species, allele D gives a dark coat and allele d (recessive) a pale coat. Researchers released marked mice into owl-proof and open enclosures on two backgrounds and recorded survival over one month. In the owl-proof enclosures, more than 95% of mice of both colours survived on both backgrounds. Results for the open enclosures:
| Background | Dark mice surviving | Pale mice surviving |
|---|---|---|
| Pale sand | 20% | 70% |
| Dark soil | 65% | 25% |
Which conclusion is best supported?
- (A) Allele D is adaptive on dark soil and harmful on pale sand, because owls catch mice that contrast with the background more often.
- (B) Allele D is harmful in general, because the mean survival of dark mice (42.5%) is lower than that of pale mice (47.5%).
- (C) Living on pale sand causes mice to develop the d allele, which protects them from owls.
- (D) Coat colour does not affect survival, because both colours survive well on at least one background.
Answer and explanation
Answer: (A). On dark soil, dark mice survive better (65% against 25%); on pale sand the pattern reverses (20% against 70%). The value of allele D depends on the environment. The owl-proof enclosures are a control, linking the differences to predation.
- (B) averages across two different environments. No mouse lives on an “average” background, so the mean hides the real pattern.
- (C) suggests the environment causes the allele. Selection acts on alleles that already exist; it does not create them.
- (D) ignores the large differences in survival within each background.
Question 4 (data analysis · core)
A fictional island lizard population crashed after rats were introduced. In 2006, conservationists released 20 lizards from a large mainland population onto the island. Researchers measured population size, mean heterozygosity across 12 genes, and the percentage of eggs that hatched.
| Year | 1990 | 1995 | 2000 | 2005 | 2010 | 2015 |
|---|---|---|---|---|---|---|
| Population size | 1400 | 260 | 140 | 120 | 230 | 410 |
| Mean heterozygosity | 0.34 | 0.21 | 0.13 | 0.12 | 0.29 | 0.31 |
| Eggs hatching / % | 82 | 63 | 44 | 41 | 71 | 78 |
(a) Describe how heterozygosity and hatching success changed between 1990 and 2005, using numbers. (b) Calculate the percentage change in hatching success from 2005 to 2015. (c) Make a claim about the effect of the 2006 release, and support it with evidence and reasoning. (d) Suggest one other explanation for the recovery after 2006, and describe how researchers could test between the two explanations.
Worked solution and suggested Marlbridge rubric
(a) As the population fell from 1400 to 120, heterozygosity fell from 0.34 to 0.12 (a drop of about 65%) and hatching success fell from 82% to 41% (41 percentage points).
(b) (78 − 41) ÷ 41 × 100 = 37 ÷ 41 × 100 = +90% (90.2%).
(c) Claim: the release of mainland lizards restored genetic diversity and improved breeding success (genetic rescue). Evidence: heterozygosity rose from 0.12 in 2005 to 0.29 in 2010, more than doubling, and hatching rose from 41% to 71% over the same period. Reasoning: the mainland lizards brought alleles the island population had lost (gene flow). Fewer offspring were homozygous for harmful recessive alleles, so more eggs developed and hatched. The larger, more varied population is more likely to contain individuals able to survive future pressures.
(d) Other explanations (any one): rat numbers fell at the same time; better weather or more food after 2006. Test: compare a similar island with the same rat control but no released lizards, or compare hatching of eggs from island × mainland and island × island crosses kept in the same conditions.
| Point | What earns it |
|---|---|
| 1 | Both falls described with at least two quoted values |
| 1 | +90% (accept 90.2%) with working |
| 1 | Claim plus evidence quoting the change in heterozygosity and hatching after 2006 |
| 1 | Reasoning via gene flow adding alleles / fewer homozygous harmful alleles |
| 1 | Plausible alternative and a test that separates it from genetic rescue |
The data show a correlation in time; do not award the last point for “it proves genetic rescue worked”.
Question 5 (constructed response · core)
A student grows one species of bacterium in two flasks. Flask 1 is started from a single cell. Flask 2 is started from a mixture of 50 strains of the same species collected from different hospitals. After both flasks reach the same cell density, she adds the same dose of an antibiotic to each, and measures cell density again after 48 hours.
(a) State the null hypothesis for this experiment. (b) Predict which flask is more likely to show regrowth after 48 hours, and justify your prediction. (c) Explain why the antibiotic is described as a selective pressure, and what would happen to the frequency of a resistance allele in a flask where regrowth occurs.
Model answer and suggested Marlbridge rubric
(a) “The genetic diversity of the starting population has no effect on cell density 48 hours after the antibiotic is added” (or: there is no difference in regrowth between flasks 1 and 2).
(b) Flask 2. Fifty strains carry many more alleles than the descendants of one cell, so it is more likely that at least one strain already carries an allele that lets it survive this antibiotic. Those cells keep dividing and the culture regrows. In flask 1 almost all cells share the founding cell’s alleles; if it was susceptible, nearly all its descendants are too. (A random resistance mutation could still arise in flask 1, but this is less likely.)
(c) The antibiotic kills susceptible cells but not resistant ones, so it changes which cells survive and reproduce: that is selection. The resistance allele was not caused by the drug. Its frequency rises sharply, because the regrown culture is made almost entirely of descendants of resistant cells.
| Point | What earns it |
|---|---|
| 1 | A correctly worded null hypothesis (no effect / no difference) |
| 1 | Flask 2 predicted, with the reason that a diverse population is more likely to contain resistant individuals |
| 1 | Explains why flask 1 is at risk (cells nearly identical) |
| 1 | Selective pressure explained and resistance allele frequency increases |
Question 6 (calculation · stretch)
Two strains of a fictional bacterium are grown together without any antibiotic. The susceptible strain S divides every 25 minutes. The resistant strain R carries a resistance allele that slows growth, so it divides every 30 minutes. The culture starts with 1000 cells of each strain, and nutrients do not run out during the experiment.
(a) Calculate the number of S cells and R cells after 300 minutes. (b) Calculate the fraction of the culture that is strain R after 300 minutes. (c) Use your answers to explain why the resistance allele can be described as harmful in this environment. (d) The antibiotic is then added and kills all S cells. Explain why a population that keeps a few R cells, even at a cost, is more resilient.
Worked solution and suggested Marlbridge rubric
(a) S: 300 ÷ 25 = 12 divisions, so 1000 × 2¹² = 1000 × 4096 = 4 096 000 cells. R: 300 ÷ 30 = 10 divisions, so 1000 × 2¹⁰ = 1000 × 1024 = 1 024 000 cells.
(b) R fraction = 1 024 000 ÷ (1 024 000 + 4 096 000) = 1 024 000 ÷ 5 120 000 = 0.20 (20%), down from 0.50 at the start.
(c) With no antibiotic, the resistance gives no benefit, but the slower division means R leaves fewer descendants (S outnumbers R by 4 to 1 after 300 minutes). Its frequency falls, so in this environment the allele is deleterious.
(d) Only R cells survive the drug, so the population recovers from them; without any R cells it would die out. An allele that is costly in one environment saves the population in another, so keeping variation protects against changes that cannot be predicted.
| Point | What earns it |
|---|---|
| 1 | Both cell numbers correct (12 and 10 divisions) |
| 1 | 0.20 (20%) |
| 1 | Allele harmful because fewer descendants without the drug |
| 1 | Resilience linked to survivors already present when the pressure changes |
Common error: using 300 ÷ 25 = 12 as the number of cells instead of the number of doublings.
Question 7 (constructed response · stretch)
Only 40 birds of a fictional island parrot remain, all in one wet mountain valley. A zoo proposes to breed from the six healthiest birds and release their offspring on a nearby island that is much drier.
(a) Identify two features of the plan that reduce the genetic diversity of the new population, and explain why each matters. (b) Explain why alleles common in the valley parrots might be harmful on the new island. (c) Suggest one change to the plan that would make the new population more resilient, and justify it.
Model answer and suggested Marlbridge rubric
(a) (1) Only six founders carry a small sample of the 40 birds’ alleles, so alleles are lost and later generations become inbred. (2) Choosing the healthiest birds selects traits that suit present conditions; less fit-looking birds may carry alleles that would help under a different pressure, such as a new disease. (Also accept: all 40 come from one valley.)
(b) The valley parrots’ alleles were favoured by selection in a wet environment. On a dry island the selective pressures differ (for example, water loss or food plants), so valley-adapted alleles could be neutral or deleterious there.
(c) Breed from as many of the 40 birds as possible, pairing unrelated birds, to keep the most alleles and limit inbreeding; or release some offspring into a habitat like the valley. More variation makes it more likely that some individuals survive future pressures.
| Point | What earns it |
|---|---|
| 1 | Small number of founders linked to loss of alleles or inbreeding |
| 1 | A second valid feature (choosing “healthiest”, single valley) with its consequence |
| 1 | Different selective pressures make valley-adapted alleles possibly harmful |
| 1 | A practical change justified by increased diversity or a better-matched environment |
How did you do?
- Q1 or Q5 wrong: re-read “Low diversity: a population at risk” and “High diversity: a resilient population” in the study guide, and look again at Figure 1.
- Q2 or Q6 calculations wrong: rework Worked example 1; check whether the question needs q, q², 2pq or 1 − q², and count doublings before multiplying.
- Q3 or Q7(b) wrong: re-read “The same allele can help or harm” and its table.
- Q4 or Q7 incomplete: re-read Worked example 2, especially the claim–evidence–reasoning steps and the limits.
Then tick off the topic checklist.