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 relationship: relative fitness = mean success (survival or offspring) of a phenotype ÷ mean success of the most successful phenotype. A calculator is assumed. All organisms, chemicals, places and data sets are fictional.
Question 1 (multiple choice · foundation)
In a fictional population of wild sunflowers, which source of variation in plant height could be acted on by natural selection to change the height of the next generation?
- (A) Plants on the sunny side of a hill grow taller than plants on the shady side.
- (B) Some plants carry an allele that makes their stems grow longer in any conditions.
- (C) Plants that were grazed early in the season grow shorter.
- (D) Plants watered by a nearby stream grow taller than plants on dry ground.
Answer and explanation
Answer: (B). Only heritable (genetic) variation is passed to offspring. An allele that lengthens stems in any conditions is inherited, so if taller plants leave more seeds, the next generation will be taller on average.
- (A), (C) and (D) all describe variation caused by the environment: light, grazing and water. The offspring of those plants will not inherit the extra or reduced height, so selection on these differences does not change the next generation.
Question 2 (multiple choice · core)
A farmer sprayed a field with a fictional herbicide, Q, every year for six years. At first it killed almost all of a weed species. By year six, most of the weeds survived spraying. Which statement best describes the role of herbicide Q?
- (A) Herbicide Q caused mutations in the weeds that made them resistant.
- (B) Herbicide Q was a selective pressure: a few weeds were already resistant, survived spraying and left most of the seeds.
- (C) Each weed gradually adjusted to herbicide Q during its life and passed this tolerance to its seeds.
- (D) This was artificial selection, because a human chose to spray the field.
Answer and explanation
Answer: (B). Resistant variants were present before spraying began. Spraying killed susceptible plants, so the resistant ones left more seeds, and resistance became common.
- (A) Mutations arise randomly; the herbicide did not create the helpful ones. It changed which plants survived.
- (C) Changes during an individual’s life are not inherited. Populations change because some individuals leave more offspring.
- (D) The farmer did not choose which weeds bred. The weeds that happened to survive did. This is natural selection in a human-changed environment; artificial selection (Topic 7.3) means humans choosing the parents.
Question 3 (multiple choice · core)
A fictional soil bacterium has a membrane pump that removes copper ions from the cell. Some cells carry one copy of the pump gene; others carry three copies. Copper at high concentration is toxic. Which prediction is best supported?
- (A) In soil polluted with copper, three-copy cells will have higher fitness, because more pump molecules remove copper faster.
- (B) In all soils, three-copy cells will have higher fitness, because more of any protein is always better.
- (C) Copy number cannot affect fitness, because the pump protein is the same in both types of cell.
- (D) In copper-polluted soil, one-copy cells will have higher fitness, because they use less energy on pumps.
Answer and explanation
Answer: (A). More gene copies mean more pump molecules. In polluted soil, cells that remove copper faster keep internal copper lower, survive and divide more. Variation in the number of molecules affects fitness in this environment.
- (B) ignores the environment. In clean soil, extra pumps cost energy and give no benefit, so three-copy cells may do worse.
- (C) Fitness can depend on how much of a protein a cell makes, not only on its type.
- (D) The energy saving is real but small compared with the cost of copper poisoning in polluted soil. This reasoning fits clean soil, not polluted soil.
Question 4 (graph · core)
A fictional freshwater fish has two forms of a digestive enzyme, A and B, produced by different alleles. Enzyme activity was measured at six temperatures (arbitrary units).
| Temperature / °C | 5 | 10 | 15 | 20 | 25 | 30 |
|---|---|---|---|---|---|---|
| Variant A activity | 40 | 62 | 80 | 70 | 45 | 20 |
| Variant B activity | 10 | 22 | 40 | 62 | 80 | 66 |
(a) State the optimum temperature for each variant. (b) Calculate how many times more active variant A is than variant B at 5 °C. (c) Over many years, the lake’s average temperature rises from about 12 °C to about 26 °C. Predict how the frequency of the variant B allele will change, and justify your prediction. (d) Explain why fish that carry one A allele and one B allele might have higher fitness than either type of homozygote in a lake whose temperature swings widely between seasons.
Worked solution and suggested Marlbridge rubric
(a) Variant A: 15 °C (activity 80). Variant B: 25 °C (activity 80).
(b) 40 ÷ 10 = 4.0 times.
(c) The B allele should become more common. At about 12 °C, A is far more active (about 69, against about 29 for B). At about 26 °C, B is more active (about 77, against about 40 for A). These values are read between the plotted points. Fish with B digest food faster in warm water, so they can grow faster and put more energy into reproduction. They leave more offspring, and the B allele spreads.
(d) A heterozygote can make both enzyme forms. In cold seasons, the A form keeps digestion working; in warm seasons, the B form does. Its overall activity is reasonably high across the whole range, so it may grow and reproduce better over a whole year than a fish that does well only in one season.
| Point | What earns it |
|---|---|
| 1 | Both optima correct and ratio 4.0 |
| 1 | B becomes more common, supported by values read from the graph at both temperatures |
| 1 | Links higher enzyme activity to growth or energy and to more offspring |
| 1 | Heterozygote makes both forms, giving activity across the seasonal range |
Accept any reasonable values read between plotted points.
Question 5 (constructed response · core)
A fungal skin disease spreads into a fictional region and kills many frogs of one species. Frogs vary in the types of antimicrobial peptides (short proteins) in their skin secretions. Frogs that make a peptide called P are much more likely to survive infection. Making P uses energy.
(a) Identify the selective pressure and state whether it is biotic or abiotic. (b) Explain how a difference in one type of molecule can lead to a difference in fitness. (c) Predict how the frequency of P-producing frogs will change over several generations in infected ponds, and state the condition that must be true for your prediction to hold. (d) Predict, with a reason, the fitness of P-producing frogs relative to non-producers in ponds the fungus never reaches.
Model answer and suggested Marlbridge rubric
(a) The fungus (a disease-causing organism). It is biotic.
(b) Peptide P is a molecule in the skin secretions that kills or slows the fungus (cell function). So frogs that make P are less likely to be infected or killed (phenotype: resistance). They are therefore more likely to survive to breeding age and leave offspring (fitness).
(c) The frequency of P-producing frogs should increase, because they survive and reproduce more than non-producers. The condition: the ability to make P must be heritable (coded by alleles passed to offspring).
(d) Their relative fitness will probably be slightly lower than that of non-producers. Without the fungus, P gives no survival benefit, but making it still costs energy that could go into growth or eggs.
| Point | What earns it |
|---|---|
| 1 | Fungus identified as the selective pressure and as biotic |
| 1 | Chain from molecule to cell or tissue function to survival and reproduction |
| 1 | Increase predicted and heritability stated as the condition |
| 1 | Lower (or equal) fitness without the fungus, explained by cost with no benefit |
For (d), accept “about equal” if the reasoning says the benefit disappears and the cost is small.
Question 6 (constructed response · stretch)
Fictional grasshoppers come in two colours, black and brown. Researchers marked 200 of each colour and released them in a recently burnt (blackened) grassland, and another 200 of each in an unburnt (brown) grassland. Two weeks later they recaptured as many marked grasshoppers as they could.
| Habitat | Colour | Released | Recaptured |
|---|---|---|---|
| Burnt | Black | 200 | 120 |
| Burnt | Brown | 200 | 60 |
| Unburnt | Black | 200 | 50 |
| Unburnt | Brown | 200 | 110 |
A student claims: “Black colour is an advantage for these grasshoppers.”
(a) Calculate the fraction recaptured for each group, and the relative fitness of the less successful colour in each habitat (use recapture as a measure of survival). (b) Evaluate the student’s claim. (c) Give one reason why recapture may not equal survival, and one reason why survival may not equal fitness.
Model answer and suggested Marlbridge rubric
(a) Burnt: black 120 ÷ 200 = 0.60; brown 60 ÷ 200 = 0.30. Unburnt: black 50 ÷ 200 = 0.25; brown 110 ÷ 200 = 0.55. Relative fitness of the less successful colour: burnt, w(brown) = 0.30 ÷ 0.60 = 0.50; unburnt, w(black) = 0.25 ÷ 0.55 = 0.45.
(b) The claim is only partly supported. Black grasshoppers did better on burnt ground (0.60 against 0.30), probably because predators such as birds find them harder to see there. But on unburnt ground black grasshoppers did worse (0.25 against 0.55). Black colour is an advantage only in the burnt habitat. Fitness depends on the environment, so the claim should say where.
(c) Recapture: some marked grasshoppers may have moved out of the study area, or been missed, rather than died. Fitness: survival over two weeks does not show how many offspring each colour produces; one colour might mate or lay eggs more successfully.
| Point | What earns it |
|---|---|
| 1 | All four fractions correct |
| 1 | Both relative fitness values (0.50 and 0.45) |
| 1 | Claim evaluated using both habitats, concluding the advantage depends on the environment |
| 1 | One valid limitation of recapture and one of survival as fitness |
Question 7 (calculation · stretch)
A fictional annual plant population has 600 plants. Each plant’s first flowering day (day of the year) is heritable. In a year when summer drought starts early, late-flowering plants run out of water before their seeds ripen.
| Flowering day | 100 | 110 | 120 | 130 | 140 |
|---|---|---|---|---|---|
| Number of plants | 50 | 150 | 200 | 150 | 50 |
| Seeds per plant (early-drought year) | 30 | 25 | 18 | 8 | 2 |
(a) Calculate the mean flowering day of the parent plants. (b) Calculate the total seeds from each class and the mean flowering day weighted by seeds (the expected mean of the offspring if each seed flowers on its parent’s day). (c) Calculate the relative fitness of day-140 plants. (d) Explain why the real change in the next generation would probably be smaller than your answer to (b) suggests.
Worked solution
(a) The distribution is symmetrical about day 120. Check: (100×50 + 110×150 + 120×200 + 130×150 + 140×50) ÷ 600 = 72 000 ÷ 600 = day 120.
(b) Seeds: 50×30 = 1500; 150×25 = 3750; 200×18 = 3600; 150×8 = 1200; 50×2 = 100. Total = 10 150. Weighted mean = (100×1500 + 110×3750 + 120×3600 + 130×1200 + 140×100) ÷ 10 150 = 1 164 500 ÷ 10 150 = day 114.7, a shift of about 5.3 days earlier.
(c) The best class is day 100 (30 seeds per plant). w(day 140) = 2 ÷ 30 = 0.067.
(d) Flowering day is only partly heritable: some of the difference between plants is caused by their surroundings (for example soil moisture or shade), and that part is not passed on. Offspring therefore do not all flower on their parent’s day, and pollen from other plants mixes alleles. So the offspring mean would move less than 5.3 days, though still earlier.
Suggested mark points (4): 1 for day 120; 1 for 10 150 seeds and day 114.7; 1 for w = 0.067; 1 for partial heritability or environmental variation explained.
Common errors: averaging the seeds-per-plant row (giving 16.6) instead of multiplying by plant numbers; dividing the weighted sum by 600 plants instead of 10 150 seeds.
How did you do?
- Q1 or Q2 wrong: re-read “Variation is the raw material” and the FAQ on artificial selection in the study guide.
- Q3, Q4 or Q5 wrong: re-read “Variation in molecules affects fitness” and practise stating the molecule → phenotype → fitness chain.
- Q6 incomplete: rework Worked example 1; a trait is favoured only in a stated environment.
- Q7 calculations wrong: check that you weighted by seeds, not plants.
Then tick off the topic checklist.