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. Data for every question: χ² = Σ (o − e)² ÷ e; degrees of freedom = number of categories − 1; critical values at p = 0.05 are 3.84 (1 df), 5.99 (2 df) and 7.81 (3 df). A calculator is assumed. All organisms named with letters, and all data sets, are fictional.
Question 1 (multiple choice · foundation)
Beetle larvae are released in a tray that is dry at one end and damp at the other. Video tracking shows that in dry air they move quickly in long, straight runs in any direction; in damp air they move slowly and turn often. After an hour, most are at the damp end. Which conclusion is best supported?
- (A) The larvae show positive taxis toward moisture, because most of them ended up at the damp end.
- (B) The larvae show kinesis: their speed and turning rate change with humidity, so they spend longer in damp air, without moving in a particular direction.
- (C) The larvae show a learned response, because they changed their behaviour during the hour.
- (D) The larvae show a physiological response, because humidity changed how much water they lost.
Answer and explanation
Answer: (B). The tracking shows no consistent direction, only changes in speed and turning. Slow movement and frequent turning keep larvae in damp air once they arrive, so they collect there. That is kinesis.
- (A) uses only the end-point. Taxis needs paths directed toward the stimulus.
- (C) is not supported: every larva reacted the same way from the start; nothing suggests experience changed the response.
- (D) What was measured is movement, a behavioural response.
Question 2 (multiple choice · core)
A student wants to test the hypothesis: “Chemicals released by leaves chewed by caterpillars attract parasitic wasps.” In a Y-shaped tube, a wasp released at the base chooses between two arms. One arm carries air from a caterpillar-chewed plant. Which source for the other arm makes the best control?
- (A) Air from a plant of the same species, age and size that has not been chewed.
- (B) Air from an empty container.
- (C) Air from a container holding caterpillars only.
- (D) Air from an unchewed plant of a different species.
Answer and explanation
Answer: (A). The hypothesis is about chewing, so the only difference between the arms should be whether the plant was chewed. Matching species, age and size keeps everything else the same.
- (B) changes two things at once: a wasp choosing the chewed plant might be attracted to any plant smell, not to chewing.
- (C) tests a different question (are wasps attracted to caterpillars?) and differs in two ways.
- (D) changes species as well as damage, so a preference could be due to the species.
Question 3 (multiple choice · core)
Sixty small crustaceans are placed in the centre of a chamber with three equal compartments: dry, medium and humid. After 20 minutes the counts are dry 12, medium 18, humid 30. Which statement is correct for the null hypothesis “the animals show no preference”?
- (A) χ² = 8.4 with 2 degrees of freedom; reject the null hypothesis.
- (B) χ² = 168 with 2 degrees of freedom; reject the null hypothesis.
- (C) χ² = 8.4; fail to reject the null hypothesis, because some animals were found in every compartment.
- (D) χ² = 8.4 with 59 degrees of freedom; fail to reject the null hypothesis.
Answer and explanation
Answer: (A). Expected = 60 ÷ 3 = 20 per compartment. χ² = (12 − 20)² ÷ 20 + (18 − 20)² ÷ 20 + (30 − 20)² ÷ 20 = (64 + 4 + 100) ÷ 20 = 168 ÷ 20 = 8.4. df = 3 − 1 = 2, critical value 5.99. 8.4 > 5.99, so reject: the spread is unlikely to be chance alone.
- (B) forgets to divide by the expected count.
- (C) uses a wrong rule: the decision comes from comparing χ² with the critical value.
- (D) uses number of animals − 1; df depends on the number of categories (3).
Question 4 (constructed response · core)
A student thinks that flowering in plant species K is controlled by night length. She grows 20 young plants from one batch of seed in each of four identical growth cabinets for six weeks, and records how many flower.
| Cabinet | Light regime (24-hour cycle) | Plants flowering |
|---|---|---|
| A | 16 h light, 8 h dark | 0 of 20 |
| B | 8 h light, 16 h dark | 19 of 20 |
| C | as B, but a 10-minute light flash in the middle of the dark period | 1 of 20 |
| D | as B, but a 10-minute dark break in the middle of the light period | 18 of 20 |
(a) Identify the independent and dependent variables. (b) State which cabinet is the control for cabinet C, and justify your choice. (c) Give two controlled variables and explain why each must be kept the same. (d) Calculate the percentage of plants flowering in B, C and D, and state what the results show about night length.
Model answer and suggested Marlbridge rubric
(a) Independent variable: the light regime, in particular the length of uninterrupted darkness. Dependent variable: the number (or percentage) of plants that flower within six weeks.
(b) Cabinet B. It has the same light and dark hours as C; the only difference is the flash. So any difference between B and C is caused by breaking up the dark period.
(c) Any two, with a reason: temperature; light intensity during the light period; water and nutrients; age and genetic source of the plants. Each could change flowering on its own, so it would confound the effect of the light regime.
(d) B: 19 ÷ 20 × 100 = 95%. C: 1 ÷ 20 × 100 = 5%. D: 18 ÷ 20 × 100 = 90%. Breaking the night stops flowering (95% → 5%), but breaking the day has almost no effect (95% → 90%). So species K responds to the length of continuous darkness: it flowers when nights are long, like a short-day plant.
| Point | What earns it |
|---|---|
| 1 | Independent and dependent variables both correct |
| 1 | B chosen and justified: it differs from C only in the flash |
| 1 | Two controlled variables, each with a reason |
| 1 | Three percentages correct |
| 1 | Conclusion uses both C and D to show that continuous darkness is what matters |
Question 5 (constructed response · core)
In a fictional burrowing rodent, adult females sometimes give a loud alarm call when a hawk appears. Observers recorded 150 hawk appearances near individual females.
| Female’s neighbours | Hawk appearances | Female called |
|---|---|---|
| close relatives nearby | 80 | 62 |
| no close relatives nearby | 70 | 14 |
Callers were caught on 5 of their 76 calls. Females that stayed silent were caught on 1 of their 74 silent occasions.
(a) Calculate the percentage of appearances in which a female called, for each group. (b) Use the capture data to state the cost of calling to the caller. (c) Explain how alarm calling can be favoured by natural selection despite this cost. (d) Explain how alarm calling could affect the survival of the population.
Model answer and suggested Marlbridge rubric
(a) With relatives: 62 ÷ 80 × 100 = 77.5%. Without relatives: 14 ÷ 70 × 100 = 20.0%.
(b) Callers were caught on 5 ÷ 76 = 6.6% of calls, against 1 ÷ 74 = 1.4% when silent. Calling draws the hawk’s attention and raises the caller’s risk about fivefold.
(c) Relatives share many of the caller’s alleles. When a call lets several relatives escape, copies of the alleles behind calling survive in them, even if the caller is sometimes caught: kin selection. The data fit this: females call far more when relatives are near.
(d) Warned animals escape more often, so more individuals survive to breed, which helps the population persist.
| Point | What earns it |
|---|---|
| 1 | Both percentages correct |
| 1 | Cost stated with the capture percentages |
| 1 | Kin selection: relatives share alleles and the data show calling is linked to relatives |
| 1 | Population-level effect linked to more individuals surviving to reproduce |
Question 6 (constructed response · stretch)
During one breeding season, researchers recorded the calling rate of six males of a fictional tree frog and how many times each mated.
| Male | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Calls per minute | 20 | 28 | 35 | 41 | 50 | 58 |
| Matings | 1 | 1 | 2 | 3 | 4 | 6 |
(a) Calculate the percentage of all matings achieved by the two fastest-calling males. (b) Explain how this signal could lead to differential reproductive success and to evolution. (c) The data are correlational. Design a playback experiment to test whether females prefer faster calls. Identify the independent variable, the dependent variable, and two controls, and justify one of them.
Model answer and suggested Marlbridge rubric
(a) Total matings = 1 + 1 + 2 + 3 + 4 + 6 = 17. Males 5 and 6: 4 + 6 = 10. 10 ÷ 17 × 100 = 58.8%.
(b) Calling is an audible signal that changes female behaviour, so faster callers father more offspring. If calling rate is partly heritable, alleles linked to fast calling become more common in the next generation: the population evolves by natural selection.
(c) Place two speakers at equal distances from a release point. Play the same recorded call from each, edited so that only the calling rate differs (for example 30 and 55 calls per minute). Release one female at a time and record which speaker she approaches first.
- Independent variable: call rate played. Dependent variable: number of females choosing each speaker.
- Controls (any two): same recording, volume and pitch from both speakers; swap sides between trials; trials with both speakers at the same rate to check for side bias; each female tested once.
- Justification example: if females split evenly in the same-rate trials, a preference in the main trials can be put down to call rate. A chi-square test against 50 : 50 then shows whether the preference is unlikely to be chance.
| Point | What earns it |
|---|---|
| 1 | 58.8% with working |
| 1 | Signal → female choice → more offspring for fast callers |
| 1 | Heritability → allele frequency change (evolution) |
| 1 | Correct independent and dependent variables, with only call rate varied |
| 1 | Two valid controls, one justified |
Accept other valid designs if only call rate differs.
Question 7 (constructed response · stretch)
Four groups of potted plants of one species were placed in a large cage with parasitic wasps that lay eggs in caterpillars. Over three days, observers counted wasp landings.
| Plant treatment | Wasp landings |
|---|---|
| undamaged | 6 |
| leaves cut with scissors | 9 |
| leaves chewed by caterpillars (caterpillars then removed) | 31 |
| leaves cut with scissors, then caterpillar saliva applied | 27 |
(a) Calculate the percentage of all landings on caterpillar-chewed plants. (b) Explain the purpose of the “cut with scissors” treatment. (c) What do the results suggest triggers the plant’s response? Use the data. (d) Explain how this response could increase the plant’s fitness.
Model answer and suggested Marlbridge rubric
(a) Total = 6 + 9 + 31 + 27 = 73. 31 ÷ 73 × 100 = 42.5%.
(b) It is a control for damage alone: it wounds the leaf without anything from the caterpillar, separating the effect of wounding from the effect of the caterpillar itself.
(c) Wounding alone attracted few more wasps than undamaged plants (9 against 6). Chewed plants attracted 31, and scissor cuts plus saliva attracted 27, three times the scissor-only count (27 ÷ 9 = 3.0). So a substance in caterpillar saliva, together with wounding, appears to trigger the plant to release attractive chemicals.
(d) Wasps lay eggs in caterpillars, which then die. Less leaf loss means more photosynthesis, growth and seeds, so plants with this response leave more offspring and the response can be favoured by natural selection.
| Point | What earns it |
|---|---|
| 1 | 42.5% |
| 1 | Scissor treatment explained as a control for wounding |
| 1 | Conclusion supported by comparing at least two pairs of numbers |
| 1 | Fitness link: fewer herbivores → more photosynthesis or seeds |
A stronger design would test each plant in a separate cage, so that the plants do not affect one another.
How did you do?
- Q1 wrong: re-read “Taxis and kinesis” and Figure 1 in the study guide.
- Q2 or Q4 wrong: rework Worked example 1, part (b): a control changes only the factor being tested.
- Q3 wrong: rework Worked example 1, part (c); check degrees of freedom and that you divided by e.
- Q5–Q7 incomplete: re-read “Communication” and “Behaviour, fitness and natural selection”, then Worked example 2.
Then tick off the topic checklist.