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 data sets and the species in Questions 3 and 5 are fictional. Data you may need: standard error of the mean, SE = s ÷ √n, where s is the sample standard deviation and n is the sample size. A calculator is assumed. Keep full calculator values until the last step, then round to the precision of the data.
Question 1 (multiple choice · foundation)
Which statement best describes phenotypic plasticity?
- (A) The ability of a single genotype to produce different phenotypes when it develops in different environments.
- (B) The ability of an organism to change its alleles so that they suit its environment.
- (C) The appearance of a new phenotype in a population after a mutation.
- (D) The way a dominant allele hides the phenotype of a recessive allele.
Answer and explanation
Answer: (A). The genotype stays the same while the phenotype changes with the environment, usually through gene expression.
- (B) is wrong because organisms cannot change their alleles to suit conditions. The environment changes which genes are expressed, not the DNA sequence.
- (C) describes mutation, which changes the genotype. Plasticity needs no change in genotype.
- (D) describes dominance, a relationship between two alleles of one gene. It has nothing to do with the environment.
Question 2 (multiple choice · core)
A gardener has a hydrangea with blue flowers. She roots several cuttings from it and plants them in a bed with alkaline soil. The next year, all the cuttings have pink flowers. Which explanation is best supported?
- (A) The alkaline soil caused a mutation in the gene for the petal pigment.
- (B) The cuttings have the same genotype as the parent; in alkaline soil, little aluminium is available to bind the pigment, so the flowers are pink.
- (C) Alkaline soil killed the plants with blue alleles, so only plants with pink alleles survived.
- (D) The pink allele is dominant, so it is expressed once the plant is moved.
Answer and explanation
Answer: (B). Cuttings are clones, so the genotype has not changed. The colour depends on soil chemistry: in acidic soil, aluminium ions are available, are taken up and turn the pigment blue; in alkaline soil they are not, and the flowers are pink. This is one genotype giving two phenotypes.
- (A) Soil pH does not rewrite DNA, and the same mutation would not appear in every cutting at once.
- (C) describes selection, but all cuttings share one genotype, so there was nothing to select between.
- (D) misuses dominance, which describes two alleles in one individual; moving a plant does not change it.
Question 3 (multiple choice · core)
Voles of a fictional species living on a cold mountain have thicker fur than voles of the same species in a warm valley. A biologist wants to find out whether the difference is mainly genetic or mainly environmental. Which approach would best answer the question?
- (A) Measure fur thickness in many more voles from each site.
- (B) Show that fur thickness is correlated with mean air temperature at ten sites.
- (C) Raise young from both populations together under identical conditions, and compare their fur when they are adult.
- (D) Breed mountain voles only with other mountain voles for several generations.
Answer and explanation
Answer: (C). This common-garden design holds the environment constant while genetic origin varies. If the difference remains, it is mainly genetic; if it disappears, it was mainly environmental (plasticity).
- (A) A larger sample makes the difference more certain, but genotype and environment still vary together.
- (B) The correlation would appear whether the cause is genetic or plastic, so it cannot tell them apart.
- (D) changes nothing, so there is nothing to compare.
Question 4 (graph · core)
Water fleas (Daphnia) can reproduce asexually, so a single female gives a clone of genetically identical offspring. Some clones grow a taller head crest when the water contains chemicals released by predators. A student rears young from three clones with and without these chemicals. Mean crest heights:
| Clone | No predator chemicals / µm | Predator chemicals / µm |
|---|---|---|
| 1 | 40 | 70 |
| 2 | 45 | 50 |
| 3 | 30 | 85 |
(a) Calculate the change in crest height, and the percentage increase, for each clone. (b) Identify the most plastic clone and the least plastic clone, using the graph. (c) Explain why the student cannot say which clone “has the tallest crest” without naming the environment. (d) Explain why using clones makes this a fair test of the effect of predator chemicals.
Worked solution and suggested Marlbridge rubric
(a)
| Clone | Change / µm | Percentage increase |
|---|---|---|
| 1 | 70 − 40 = 30 | 30 ÷ 40 × 100 = 75% |
| 2 | 50 − 45 = 5 | 5 ÷ 45 × 100 = 11% |
| 3 | 85 − 30 = 55 | 55 ÷ 30 × 100 = 183% |
(b) Clone 3 is the most plastic: steepest line, largest change. Clone 2 is the least plastic: its line is nearly flat.
(c) The lines have different slopes and cross. Without predator chemicals clone 2 is tallest (45 µm); with them clone 3 is (85 µm). This genotype-by-environment interaction means the ranking depends on the environment.
(d) Within a clone, all individuals have the same genotype. So a difference between treatments for one clone is caused by the predator chemicals, not by different alleles.
| Point | What earns it |
|---|---|
| 1 | All three changes and percentages correct |
| 1 | Clone 3 most and clone 2 least plastic, linked to slope |
| 1 | Ranking changes between environments, described as an interaction |
| 1 | Same genotype within a clone, so the difference is environmental |
Question 5 (constructed response · core)
A fictional northern hare grows a white coat in autumn. Is the moult triggered by shorter days or by falling temperatures?
(a) Explain why field observations alone cannot answer this. (b) Describe an experiment with captive hares to test this: groups, variable measured, variables kept constant. (c) All hares on short days turn white, whatever the temperature. State the conclusion, and explain how one genotype gives a brown summer coat and a white winter coat. (d) Suggest how this response could become harmful as snow cover becomes shorter.
Model answer and suggested Marlbridge rubric
(a) In the wild, day length and temperature fall together. They are confounded, so either could be the cause.
(b) Split hares of similar age and genetic background (for example, siblings) across four rooms: short days and cold; short days and warm; long days and cold; long days and warm. Measure the percentage of white fur weekly. Keep constant: diet, cage size, light intensity and handling.
(c) Day length, not temperature, triggers the moult. The alleles do not change between seasons. Shorter days change gene expression in the skin: new hairs grow with little pigment because melanin genes are expressed less. One genotype, two phenotypes.
(d) Day length does not change with climate, but snow cover does. A hare can turn white on schedule while the ground is still brown, so predators see it easily.
| Point | What earns it |
|---|---|
| 1 | Day length and temperature are confounded |
| 1 | Design varies the two factors independently, with a measured outcome |
| 1 | At least two sensible controlled variables, and genetically similar hares |
| 1 | Day length, and a change in gene expression with the same genotype |
| 1 | Mismatch between coat colour and snow cover, with a consequence (predation) |
Question 6 (constructed response · stretch)
Baker’s yeast has two haploid mating types, a and α. Each releases a pheromone detected by the other type. A student measures the a-pheromone released by a cells in three conditions (arbitrary units, three cultures each):
| Condition | Culture 1 | Culture 2 | Culture 3 |
|---|---|---|---|
| W: a cells alone | 9 | 11 | 10 |
| X: a cells mixed with α cells | 46 | 50 | 48 |
| Y: a cells in filtered medium in which α cells had grown (no cells) | 43 | 47 | 45 |
(a) Calculate the mean for each condition, and the fold change for X and Y compared with W. (b) Explain what the result for condition Y shows that condition X alone could not. (c) Explain, in terms of gene expression, how the presence of the opposite mating type changes the phenotype of the a cells.
Model answer and suggested Marlbridge rubric
(a) Means: W = (9 + 11 + 10) ÷ 3 = 10; X = 48; Y = 45. Fold change: X = 48 ÷ 10 = 4.8; Y = 45 ÷ 10 = 4.5.
(b) In X, the a cells could be responding to contact with α cells or to something they release. Y has no α cells, yet the response is almost as large, so the signal is a released chemical (α-pheromone); contact is not needed.
(c) α-pheromone binds a surface receptor on the a cell. A signalling pathway inside the cell switches on mating genes, including the a-pheromone genes, so the cell releases more. The genotype is the same in all three conditions; only the presence of the other mating type, and so the phenotype, has changed.
| Point | What earns it |
|---|---|
| 1 | All three means and both fold changes correct |
| 1 | Y shows the signal is a released chemical, not contact |
| 1 | Receptor binding leads to a change in gene expression (mating genes on) |
| 1 | Genotype unchanged, so the difference is environmental |
Question 7 (calculation · stretch)
In a fictional study, six adults had two small skin patches measured with a meter that gives a melanin index (higher = more melanin). One patch received low doses of UV light over two weeks; the other was kept covered.
| Volunteer | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| UV patch, before | 30 | 28 | 35 | 32 | 26 | 29 |
| UV patch, after | 41 | 36 | 47 | 42 | 33 | 40 |
| Covered patch, before | 30 | 28 | 35 | 32 | 26 | 29 |
| Covered patch, after | 30 | 29 | 35 | 31 | 26 | 30 |
(a) Calculate the change for each volunteer on each patch, and the mean change for each patch. (b) Calculate the percentage increase in the mean melanin index of the UV patch. (c) The sample standard deviation of the UV-patch changes is 1.94. Calculate the standard error, and use ±2 SE to judge whether the increase is likely to be real. (For the covered patch, the mean change ± 2 SE is −0.4 to 0.8.) (d) Explain why each person serves as their own control, and why the covered patch is needed. (e) A volunteer asks whether her future children will be born with darker skin because of the study. Answer her, using the idea of gene expression.
Worked solution
(a) UV patch changes: 11, 8, 12, 10, 7, 11. Sum = 59; mean = 59 ÷ 6 = 9.8. Covered patch changes: 0, 1, 0, −1, 0, 1. Sum = 1; mean = 1 ÷ 6 = 0.2.
(b) Mean before = 180 ÷ 6 = 30.0; mean after = 239 ÷ 6 = 39.8. Increase = (39.83 − 30.0) ÷ 30.0 × 100 = 33% (32.8%). Dividing by the “after” value (24.7%) is a common error.
(c) SE = s ÷ √n = 1.94 ÷ √6 = 1.94 ÷ 2.449 = 0.79. Mean ± 2 SE = 9.83 ± 1.58, so about 8.2 to 11.4. This is far from zero and does not overlap the covered patch’s interval (−0.4 to 0.8), so the increase is very unlikely to be chance alone.
(d) Two patches on one person remove differences between people, such as genotype and normal skin colour. The covered patch shows what happens without UV (for example, seasonal change or meter drift); it barely changed.
(e) No. UV changed gene expression in pigment cells of the exposed skin, not the alleles in her eggs. Her children inherit her genotype, including the ability to tan, not the tan.
Suggested mark points (5): 1 for both mean changes; 1 for about 33% using the “before” mean as the base; 1 for SE ≈ 0.79 and the interval 8.2–11.4; 1 for a conclusion from non-overlapping intervals and a reason for the covered patch; 1 for explaining that the change is in gene expression in body cells, not in the gametes’ DNA.
How did you do?
- Q1 or Q2 wrong: re-read “How the environment reaches the genes” and the examples table in the study guide.
- Q3 or Q5 incomplete: re-read Worked example 1(c) on why clones and controlled conditions separate genotype from environment.
- Q4 wrong: re-read “Representing plasticity: reaction norms” and look again at Figure 1.
- Q6 or Q7 incomplete: name the signal, the change in gene expression, and that the genotype is unchanged. For percentages divide by the starting value; for SE use √n.
Then tick off the topic checklist.