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 relationships: mean offspring per individual = total offspring ÷ number of individuals (or fraction surviving × offspring per survivor); relative fitness = mean offspring of a phenotype ÷ mean offspring of the most successful phenotype. A calculator is assumed. All organisms, places and data sets are fictional.
Question 1 (multiple choice · foundation)
Which set of conditions must all be present for natural selection to change a population?
- (A) Individuals vary in a trait; the variation is heritable; individuals with different versions of the trait leave different numbers of offspring.
- (B) Individuals change their traits to suit the environment; these changes are passed to offspring; the population grows.
- (C) The environment stays constant; all offspring survive; individuals vary in a trait.
- (D) A new mutation appears in every individual at the same time; the mutation is helpful; resources are unlimited.
Answer and explanation
Answer: (A). Selection needs variation to choose between, inheritance so the result carries to the next generation, and differences in reproductive success so some variants leave more offspring.
- (B) describes inheritance of characteristics acquired during life. Individuals cannot change their genes to suit the environment, and acquired changes are not inherited.
- (C) If all offspring survive there is no differential survival, so no selection. A constant environment is not required.
- (D) Mutations arise randomly in individuals, not in everyone at once; and with unlimited resources there is little competition.
Question 2 (multiple choice · core)
Four female birds in a fictional population were followed for life.
| Bird | Lifespan / years | Offspring that survived to breed |
|---|---|---|
| A | 9 | 2 |
| B | 3 | 6 |
| C | 5 | 4 |
| D | 7 | 0 |
Which bird had the highest evolutionary fitness?
- (A) Bird A, because she lived longest.
- (B) Bird B, because she left the most offspring that survived to breed.
- (C) Bird C, because she balanced a long life with many offspring.
- (D) Bird D, because she lived long without spending energy on breeding.
Answer and explanation
Answer: (B). Fitness is measured by reproductive success. Bird B left 6 offspring that reached breeding age, more than any other bird, even though she lived only 3 years.
- (A) treats lifespan as fitness. Bird A left only 2 breeding offspring.
- (C) There is no extra credit for balance; Bird C left 4, fewer than Bird B.
- (D) Bird D left no breeding offspring, so her fitness is zero, however long she lived.
Question 3 (multiple choice · core)
In a fictional moth population, pale moths leave a mean of 1.2 offspring each, intermediate moths 2.4 and dark moths 3.0. What is the relative fitness of the pale moths?
- (A) 0.40
- (B) 0.50
- (C) 0.60
- (D) 2.5
Answer and explanation
Answer: (A). The most successful phenotype is dark (3.0). w(pale) = 1.2 ÷ 3.0 = 0.40.
- (B) divides by the intermediate phenotype (1.2 ÷ 2.4). Relative fitness is measured against the most successful phenotype.
- (C) is 1 − 0.40. That is the proportional shortfall, not the relative fitness.
- (D) divides the wrong way (3.0 ÷ 1.2). Relative fitness can never be above 1 for a phenotype that is not the best.
Question 4 (graph · core)
A fictional pond contains a snail species that lives for one year, so each year is one generation. Crayfish, which crush snail shells, were common in years 1–4 but died out from disease at the end of year 4. Thick shells resist crushing but take more energy to build, which leaves less energy for making eggs. The graph shows the mean shell thickness of adult snails each year.
| Year | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| Mean shell thickness / mm | 0.40 | 0.43 | 0.46 | 0.49 | 0.47 | 0.45 | 0.43 | 0.41 |
(a) Describe the graph: state what each axis shows and describe the trend in each period, with the rate of change per year. (b) Explain the change in years 1–4 using natural selection. (c) Explain why the direction of change reversed after year 4. (d) State one piece of extra evidence you would need to be sure the change is evolution and not just a response of each snail to its surroundings.
Worked solution and suggested Marlbridge rubric
(a) The x-axis is time in years (one generation per year); the y-axis is mean shell thickness in mm. From year 1 to year 4, thickness rises steadily from 0.40 to 0.49 mm, an increase of 0.09 mm, or 0.03 mm per year. From year 4 to year 8 it falls steadily from 0.49 to 0.41 mm, a decrease of 0.08 mm, or 0.02 mm per year. The dashed line marks the loss of the crayfish.
(b) Snails vary in shell thickness. While crayfish were common (a biotic selective pressure), thick-shelled snails were less likely to be crushed, so more of them survived to lay eggs. If thickness is heritable, their offspring also had thick shells, so the mean rose each generation.
(c) After the crayfish died out, thick shells no longer gave a survival advantage, but they still cost energy. Thin-shelled snails could put more energy into eggs, so they left more offspring and had higher fitness. Selection now favoured thinner shells, so the mean fell. The environment changed, so the direction of selection changed.
(d) Any one: evidence that shell thickness is inherited (for example, raising offspring of thick- and thin-shelled parents in the same conditions and finding they differ); counts of offspring per snail for each shell type; evidence that the change is not caused by a change in water chemistry (for example calcium) affecting every snail’s growth.
| Point | What earns it |
|---|---|
| 1 | Both axes named with units and both trends with correct rates (0.03 and 0.02 mm per year) |
| 1 | Years 1–4: variation, crayfish as selective pressure, higher survival and reproduction of thick-shelled snails, inheritance |
| 1 | After year 4: cost of thick shells, thin-shelled snails leave more offspring, so direction reverses |
| 1 | A valid piece of extra evidence linked to heritability or reproductive success |
Do not award the second point for “snails grew thicker shells to protect themselves”: that describes individuals changing, not selection.
Question 5 (constructed response · core)
A fictional lake contains a population of small fish whose body colour ranges from pale silver to dark grey. A large wading bird that hunts in the shallows, where the lake bed is pale sand, arrives and stays.
(a) Using each step of Darwin’s argument, explain how the colour of the fish population could change over many generations. (b) Identify the selective factor in this example and state whether it is biotic or abiotic. (c) Explain why a single fish cannot become paler in response to the birds and pass this on.
Model answer and suggested Marlbridge rubric
(a) The fish produce more young than the lake can support, so they compete for food and safe places. They vary in colour, and colour is (assume) heritable. Against pale sand, dark fish are easier for the birds to see, so pale fish are more likely to survive to breed. Pale fish therefore leave more offspring, and these offspring tend to be pale. Over many generations, the proportion of pale fish rises and the mean colour of the population becomes paler.
(b) The wading bird, a predator. It is a biotic factor. (The pale sand is abiotic, but it is the predator’s hunting that does the selecting; sand alone would not change survival.)
(c) A fish’s colour is set by its genes. Being hunted does not change those genes, and any change during its life (if there were one) would not be in its eggs or sperm. The population changes because pale fish leave more offspring, not because any fish changes.
| Point | What earns it |
|---|---|
| 1 | Overproduction and competition or limited resources |
| 1 | Variation in colour and heritability |
| 1 | Pale fish survive and reproduce more, so pale becomes more common over generations |
| 1 | Predator identified as biotic and (c) explains that individuals do not change their heritable traits |
Accept an answer that argues the sand’s colour is the abiotic condition that makes the predator’s choice matter, provided the predator is named as the agent of selection.
Question 6 (constructed response · stretch)
A fictional lizard population has two colour patterns. Equal numbers (400 of each) were followed through one season.
| Pattern | Starting adults | Fraction surviving to breed | Offspring per survivor (reaching adulthood) |
|---|---|---|---|
| Striped | 400 | 0.80 | 1.5 |
| Plain | 400 | 0.50 | 3.0 |
A student concludes: “Striped lizards are better adapted, because they survive better. The population will become mostly striped.”
(a) Calculate the mean offspring per starting adult and the relative fitness of each pattern. (b) Calculate the fraction of plain lizards in the next generation, assuming pattern is inherited exactly. (c) Evaluate the student’s conclusion.
Model answer and suggested Marlbridge rubric
(a) Striped: 0.80 × 1.5 = 1.2 offspring per adult. Plain: 0.50 × 3.0 = 1.5. The best phenotype is plain, so w(plain) = 1.00 and w(striped) = 1.2 ÷ 1.5 = 0.80.
(b) Striped offspring = 400 × 1.2 = 480. Plain offspring = 400 × 1.5 = 600. Fraction plain = 600 ÷ (480 + 600) = 600 ÷ 1080 = 0.56 (up from 0.50).
(c) The conclusion is not supported. Striped lizards do survive better (0.80 against 0.50), but fitness is measured by reproductive success, and plain lizards leave more offspring per individual. The data predict that plain lizards will become more common, not less. The conclusion also rests on one season; if conditions change, the result could differ, and real inheritance is not exact.
| Point | What earns it |
|---|---|
| 1 | Both means (1.2 and 1.5) |
| 1 | Both relative fitness values, dividing by the plain (best) phenotype |
| 1 | Fraction plain ≈ 0.56 with working |
| 1 | Rejects the claim because fitness is offspring, not survival, and gives a limitation (one season, inheritance assumed, environment may change) |
Question 7 (calculation · stretch)
Two strains of a fictional bacterium, C and D, grow together in a culture. Bacteria reproduce asexually, so each cell’s offspring belong to its own strain. At 15 °C, strain C is the fitter strain: for every 1.0 offspring cell left by a C cell, a D cell leaves 0.80 (w(D) = 0.80). At the start, 10% of cells are strain C.
(a) Calculate the percentage of strain C after one generation and after two generations at 15 °C. (b) The culture is then moved to 30 °C, where strain D is fitter and w(C) = 0.70. Calculate the percentage of strain C after one generation at 30 °C. (c) Use your answers to explain how a fluctuating abiotic environment affects the rate and direction of evolution.
Worked solution
(a) Work per 1000 cells.
- Generation 0: 100 C and 900 D. Offspring: C = 100 × 1.0 = 100; D = 900 × 0.80 = 720. Fraction C = 100 ÷ 820 = 0.122, so 12.2%.
- Generation 1 (per 1000): 122 C and 878 D. Offspring: C = 122 × 1.0 = 122; D = 878 × 0.80 = 702.4. Fraction C = 122 ÷ 824.4 = 0.148, so 14.8%.
(b) Generation 2 (per 1000): 148 C and 852 D. Offspring: C = 148 × 0.70 = 103.6; D = 852 × 1.0 = 852. Fraction C = 103.6 ÷ 955.6 = 0.108, so 10.8%.
(c) At 15 °C, strain C rose by about 2–3 percentage points per generation: selection pushed the population towards C. At 30 °C, the direction reversed and C fell by 4 percentage points in one generation, because the fitness difference was larger (w = 0.70, not 0.80) and the population moved faster. If temperature keeps changing, neither strain takes over, and the population swings back and forth. The rate depends on the size of the fitness difference; the direction depends on which strain the current environment favours.
Suggested mark points (4): 1 for 12.2%; 1 for 14.8%; 1 for 10.8%; 1 for linking the reversal to the change in temperature and the faster change to the larger fitness difference.
Common errors: dividing by 1000 instead of by the total number of offspring leaves C stuck at 10.0%; multiplying C by the fitness instead of D at 15 °C reverses the direction.
How did you do?
- Q1 or Q5 wrong: re-read “Darwin’s argument, step by step” and Figure 1 in the study guide.
- Q2, Q3 or Q6 wrong: rework Worked example 1; remember that fitness counts offspring, and relative fitness divides by the best phenotype.
- Q4 incomplete: re-read Worked example 2 and “Changing environments change the direction of selection”.
- Q7 calculations wrong: check that you divided by the total number of offspring, not by the starting number.
Then tick off the topic checklist.