IB MYP Sciences – Evolution and Interactions Between Organisms Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set is for IB MYP Sciences and is aligned to the International Baccalaureate Organization, Middle Years Programme Subject Brief – Sciences, from 2014. It covers evolution and interactions between organisms, two topics the brief lists for the on-screen examinations, and suits MYP years 4 and 5. MYP has no prescribed content list – schools design their own units – so the science here is standard content for those topics. All data are fictional.

Each question is labelled with the criterion it trains. Real MYP work is judged against criterion level descriptors (eight achievement levels, 1–8, in four bands), so the [1] points below are a revision aid, not IB marks. Your teacher will share the task-specific clarifications for your own assessments.

Related: study guide · criteria in practice questions.

Questions

1. (Criterion A)

(a) Distinguish between continuous and discontinuous variation, giving one example of each. [2] (b) State one environmental cause of variation in the height of a plant species. [1]

2. (Criterion A) Name the type of symbiosis in each case. [3]

• (i) A flea feeds on a dog’s blood, and the dog’s skin becomes irritated and sore.
• (ii) In a lichen, a fungus shelters an alga and supplies water and minerals; the alga supplies sugars.
• (iii) An orchid grows on a high tree branch and gains light; the tree is not affected.

3. (Criterion A) A hospital uses the same antibiotic for many years. Explain how a population of bacteria can become resistant to it. [5]

4. (Criterion A) In a fictional pond: algae are eaten by water fleas and mayfly larvae; water fleas are eaten by dragonfly larvae and small fish; mayfly larvae are eaten by small fish; small fish are eaten by herons.

(a) Identify the producer and one secondary consumer. [2] (b) A disease kills most of the dragonfly larvae. Predict and explain the effect on the water fleas and on the small fish. [3]

5. (Criterion C) In a fictional woodland, the energy in each trophic level per year is: producers 40 000 kJ m⁻²; primary consumers 3200 kJ m⁻²; secondary consumers 384 kJ m⁻².

(a) Calculate the efficiency of energy transfer from producers to primary consumers. [2] (b) Calculate the efficiency of energy transfer from primary to secondary consumers. [2] (c) Give two reasons why most energy in the primary consumers does not reach the secondary consumers. [2]

6. (Criteria B and C) A student placed eight 0.5 m × 0.5 m quadrats at random on a 12 m × 15 m lawn and counted dandelions: 5, 8, 2, 6, 0, 7, 4, 4.

(a) Estimate the number of dandelions on the lawn. Show your working. [3] (b) Explain why the quadrats were placed at random. [1] (c) Suggest two ways to make the estimate more reliable. [2]

7. (Criterion C) In a fictional garden, 80 snails were caught and marked with a dot of paint on the shell, then released. Two days later, 75 were caught, of which 15 were marked.

(a) Estimate the snail population. [2] (b) State two assumptions this method makes. [2] (c) Rain washed the paint off some shells before the second sample. Explain the effect on the estimate. [2]

8. (Criteria B and C) Two duckweed species, A and B, float on ponds and reproduce by making new fronds.

(a) Design an investigation to find out whether species A out-competes species B. Include a testable hypothesis with a scientific reason, the variables, and how you would make the results reliable. [7] (b) In a trial, species B grew to 96 fronds alone but only 36 fronds when grown with species A. Calculate the percentage decrease. [2]

9. (Criterion C) A beetle on a fictional island has pale and dark forms. After a new factory opened, soot darkened the tree bark. Birds eat the beetles.

Generation	1	2	3	4	5
Dark beetles	40	63	120	205	279
Total beetles sampled	400	420	430	440	450

(a) Calculate the percentage of dark beetles in generations 3 and 5, to 1 decimal place. [2] (b) Describe the trend in the percentage of dark beetles. [2] (c) Explain the trend using the theory of natural selection. [4]

10. (Criterion A)

(a) Explain how the forelimbs of a human, a bat and a whale provide evidence for evolution. [2] (b) Suggest why the fossil record is incomplete. [1] (c) A protein differs by 2 amino acids between fictional species X and Y, and by 19 between X and Z. Deduce which species is most closely related to X. [1]

11. (Criteria C and D) Hares and foxes were counted on the fictional island of Norland.

Year	1	2	3	4	5	6	7	8
Hares	200	420	610	380	190	230	450	600
Foxes	12	15	24	33	26	15	13	20

(a) Describe and explain the relationship between the two populations in years 1–5. [3] (b) Norland’s farmers propose releasing a non-native predator to reduce the hares that damage their crops. Evaluate this proposal. [6]

Answers

1. (a) Continuous variation takes any value in a range, for example body mass [1]. Discontinuous variation falls into distinct groups, for example ABO blood group [1]. (b) Light intensity, or soil mineral content, or water supply [1]. [3] Examiner insight: A distinction needs a definition and an example for each type; an example alone earns nothing.

2. (i) Parasitism [1]; (ii) mutualism [1]; (iii) commensalism [1]. [3] Examiner insight: Read the stated effect on each partner – “not affected” signals commensalism, and “harmed” signals parasitism.

3. Bacteria in the population vary; a random mutation gives some an allele for resistance [1]. The antibiotic acts as a selection pressure, killing non-resistant bacteria [1]. Resistant bacteria survive and reproduce [1]. They pass the resistance allele to their offspring [1]. Over many generations, the proportion of resistant bacteria in the population increases [1]. [5] Examiner insight: “The bacteria became immune” earns no credit; describe selection of an existing mutation.

4. (a) Producer: algae [1]. Secondary consumer: dragonfly larvae or small fish [1]. (b) Water fleas increase [1], because fewer dragonfly larvae eat them [1]. Small fish may increase, because more water fleas are available as food [1]. [5] Examiner insight: “Explain” needs a reason for every predicted change; a direction on its own earns only half the credit.

5. (a) 3200 ÷ 40 000 × 100 [1] = 8.0% [1]. (b) 384 ÷ 3200 × 100 [1] = 12% [1]. (c) Any two of: energy lost as heat from respiration; energy used in movement; parts not eaten or not digested, passed out in faeces. One point each [1] [1]. [6] Examiner insight: An efficiency above 100% shows the division was upside down; always put the higher trophic level’s energy on top.

6. (a) Mean = 36 ÷ 8 = 4.5 per quadrat [1]. Density = 4.5 ÷ 0.25 = 18 per m² [1]. Estimate = 18 × (12 × 15) = 18 × 180 = 3240 dandelions [1]. (b) To avoid bias from choosing areas that look dense or sparse [1]. (c) Use more quadrats [1]; repeat the whole survey and compare the means [1]. [6] Examiner insight: Missing the divide-by-0.25 step gives 810, which loses the density point but can still earn the final point as follow-through if the method is shown.

7. (a) N = (80 × 75) ÷ 15 [1] = 400 snails [1]. (b) Any two of: marks do not affect survival; marked snails mix fully with the population; no significant births, deaths or migration between samples; marks are not lost. One point each [1] [1]. (c) Fewer marked snails would be recognised in the second sample, so the number on the bottom of the formula falls [1]; the estimate would be too high – for example, 10 recaptures gives 600 [1]. [6] Examiner insight: In part (c) state the direction of the error (overestimate) and the reason; “the result would be inaccurate” earns nothing.

8. (a)

• Hypothesis: species B will produce fewer fronds when grown with species A than alone [1], because A takes up more light and minerals from the shared water surface [1].
• Independent variable: whether B is grown alone or with A [1].
• Dependent variable: number of B fronds after a fixed time, for example 14 days [1].
• Control variables, any two of: same starting number of fronds; same volume and type of water and mineral supply; same light intensity and temperature. One point each [1] [1].
• Reliability: at least three containers of each condition, then calculate a mean [1].

(b) (96 – 36) ÷ 96 × 100 [1] = 62.5% [1]. [9] Examiner insight: A hypothesis earns its second point only with a scientific reason; “B will grow less” alone is a prediction, not a justified hypothesis.

9. (a) Generation 3: 120 ÷ 430 × 100 = 27.9% [1]. Generation 5: 279 ÷ 450 × 100 = 62.0% [1]. (b) The percentage of dark beetles rises every generation, from 10.0% to 62.0% [1]. The biggest rise is between generations 3 and 4, from 27.9% to 46.6% [1]. (c) Beetle colour varies and is inherited [1]. On soot-darkened bark, pale beetles are easier for birds to see, so more pale beetles are eaten [1]. Dark beetles are more likely to survive and reproduce [1]. They pass the allele for dark colour to their offspring, so its frequency rises over generations [1]. [8] Examiner insight: The sample size changes each generation, so the trend must be read from percentages, not raw counts.

10. (a) The limbs are homologous – the same pentadactyl bone arrangement [1] – adapted for different uses, which suggests they share a common ancestor [1]. (b) Fossils form only in rare conditions, such as rapid burial with few decomposers, so most organisms leave none [1]. (c) Species Y, because its protein differs from X’s by fewer amino acids [1]. [4] Examiner insight: “They look similar” is not enough for (a); the point comes from naming the shared underlying structure.

11. (a) Hares peak in year 3 (610), then foxes peak in year 4 (33) [1]. Foxes increase after hares because more food means more foxes survive and breed – a lag [1]. Hares then fall from 610 to 190 by year 5 as predation increases [1]. (b) Benefits:

• Less crop damage and higher yields [1].
• Fewer poisons needed [1].

Limitations:

• The predator may eat native species instead of, or as well as, hares [1].
• With no natural predators on Norland, its own population could grow unchecked [1].
• Once released, it would be very hard to remove [1].

Judgement: too risky unless trials show the predator targets hares; a native control is safer [1]. [9] Examiner insight: “Evaluate” needs both sides and a final judgement linked to them; listing only benefits earns none of the limitation or judgement points.

Where marks are usually lost

• Writing that bacteria “adapt” or “become immune” instead of describing selection.
• Leaving out inheritance: the allele must be passed to offspring.
• Describing a data trend without quoting values from the table.
• Forgetting the quadrat area (0.25 m²) when scaling to the whole lawn.
• Putting the second-sample total, not the marked recaptures, on the bottom of the capture–mark–recapture formula.
• Hypotheses with no scientific reasoning, or control variables with no stated value or method.
• Evaluations with no conclusion, or with benefits only.

Next steps

• Recap with the revision notes.
• Relearn any weak area from the study guide.
• See the MYP Sciences course hub and the printable checklist.
• Prepare for the 50-mark task with investigation skills exam preparation.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

International Baccalaureate Organization, Middle Years Programme Subject Brief – Sciences, from 2014. The brief lists evolution and interactions between organisms among the topics explored in MYP sciences on-screen examinations.