IB DP Biology: Populations and communities: sampling, growth, interactions (DP Biology) – 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.

These questions cover IB Diploma Programme Biology section C4.1 (Populations and communities), from the IB Biology guide, first assessment 2025. The content is identical for Standard Level and Higher Level. Read the study guide first, or use the revision notes. Hub: IB DP Biology. See also the checklist and the free diagnostics.

Questions

1. Define a population, and state how reproductive isolation is used to tell two populations of one species apart. [2]

2. Explain why ecologists estimate population size by random sampling instead of counting every individual, and what is meant by sampling error. [3]

3. Twelve random quadrats, each 1 m², were placed in a meadow of 2400 m². Counts of one sessile plant species were: 3, 0, 5, 2, 4, 1, 6, 3, 2, 4, 5, 1. (a) Calculate the mean number per quadrat. [1] (b) Estimate the population in the meadow. [2] (c) The standard deviation is 1.86. Explain what this indicates about how the plants are spread. [2]

4. Researchers marked 60 shrews and released them. A later sample contained 50 shrews, of which 12 were marked. (a) Calculate the Lincoln index estimate. [2] (b) State two assumptions made. [2] (c) Marks fade between samples. State the effect on the estimate. [1]

5. (a) Define carrying capacity. [1] (b) State one resource that can limit it. [1] (c) Explain how one density-dependent factor returns a population above carrying capacity towards it. [2]

6. The number of duckweed fronds in a tank was counted daily: 10, 13, 17, 22, 29. (a) Calculate the percentage increase from day 0 to day 1. [1] (b) Describe how to use a graph to test whether growth is exponential. [2] (c) Explain why this growth cannot continue indefinitely. [2]

7. Yeast cells (millions per mL) at 4-hour intervals from 0 h to 24 h were: 2, 6, 18, 46, 70, 78, 80. (a) Identify the interval of fastest growth and calculate the rate. [2] (b) Estimate the carrying capacity and explain the plateau. [2] (c) Explain why a sigmoid curve is a model, not a description of every real population. [2]

8. (a) Define mutualism. [1] (b) For legume root nodules, state a benefit to each organism. [2] (c) For zooxanthellae in hard corals, state a benefit to each organism. [2] (d) Distinguish mutualism from parasitism. [1]

9. A grass species X was introduced to a region where a native grass Y grows. Researchers removed X from some plots and left others untouched, then measured Y. (a) Outline how the experiment should be designed and what result would indicate competition. [3] (b) Explain why the result would indicate but not prove competition. [2]

10. Two snail species were surveyed at 60 pond sites. Both present: 4. Only the first present: 21. Only the second present: 19. Neither present: 16. (a) State a null hypothesis. [1] (b) Calculate the expected number of sites with both present. [2] (c) The other expected values are 15.42, 13.42 and 21.58. Calculate χ². [2] (d) Using a critical value of 3.84, state the degrees of freedom and the conclusion. [2] (e) Explain why the result does not prove competition. [2]

11. Records of snowshoe hare and Canada lynx show numbers rising and falling in cycles of about ten years. (a) Describe the pattern shown by the two populations. [2] (b) Explain this as delayed negative feedback. [2] (c) Distinguish top-down from bottom-up control, and state which the hare cycle suggests may also matter. [2]

12. (a) State one similarity between allelopathy and antibiotic secretion. [1] (b) Give one example of each. [2] (c) State which kind of organisms produce each. [1]

Answers

1. Individuals of one species found together in one area, which normally breed with each other [1]. Two populations are distinguished because little or no interbreeding occurs between them (reproductive isolation) [1]. [2] Examiner insight: Mention both the same-species, same-area idea and the breeding link; “a group of animals” alone is too vague.

2. Counting all individuals is too slow, costly or impossible, and numbers may change during the count [1]. Samples must be random to avoid bias in where you sample [1]. Sampling error is the difference between the estimate and the true population size [1]. [3] Examiner insight: A reason for estimating and a reason for randomness are separate points; give both.

3. (a) Total 36 ÷ 12 = 3.0 per quadrat [1] (b) Density 3 per m² [1]; 3 × 2400 = 7200 plants [1] (c) SD is large compared with the mean (1.86 against 3) [1], so counts vary a lot: plants are unevenly spread or clumped [1]. [5] Examiner insight: Compare the standard deviation with the mean; a number alone says nothing about spread.

4. (a) 60 × 50 ÷ 12 [1] = 250 shrews [1] (b) Marked animals mix fully with the population [1]; no births, deaths, immigration or emigration between samples [1] (marks must not harm the animals or fade is also valid) (c) Fewer marked animals are recognised, so R is too low and the estimate is too high [1]. [5] Examiner insight: Name the assumption in full. Say what would break it, as in part (c), rather than only naming “closed population”.

5. (a) The maximum population size an environment can sustain indefinitely [1] (b) Food, water, light, space or nesting sites (any one) [1] (c) Example: disease spreads more easily when density is high [1], so deaths rise and numbers fall back towards carrying capacity (negative feedback) [1]. [4] Examiner insight: A density-dependent answer should say why the effect grows with density, then link to the return to carrying capacity.

6. (a) (13 − 10) ÷ 10 × 100 = 30% [1] (b) Plot the number of fronds on a logarithmic vertical axis against time on a normal horizontal axis [1]. A straight line indicates exponential growth, as the percentage increase is constant [1] (c) Space, light and nutrients become limiting [1]; density-dependent factors reduce growth until births equal deaths [1]. [5] Examiner insight: State which axis is logarithmic, and what a straight line means.

7. (a) 8 to 12 h [1]: (46 − 18) ÷ 4 = 7 million cells per mL per hour [1] (b) About 80 million per mL [1]; births roughly equal deaths because nutrients and space are limiting [1] (c) Real populations may overshoot, oscillate or crash [1]; the model ignores variation and interacting factors, so it simplifies a complex system [1]. [6] Examiner insight: Include units in rate answers, and show the subtraction.

8. (a) An interspecific relationship benefiting both species [1] (b) Rhizobium fixes nitrogen for the plant [1]; the plant gives carbohydrate and shelter [1] (c) Zooxanthellae supply sugars to the coral [1]; the coral gives shelter and carbon dioxide [1] (d) In parasitism one species gains and its host is harmed [1]. [6] Examiner insight: Each organism needs its own benefit. A benefit for only one partner leaves the point unfinished.

9. (a) Several plots with X removed, plus control plots with X present, of similar size and conditions [1]; measure growth or density of Y in both [1]; if Y is higher where X was removed, competition is indicated [1] (b) Y might benefit from other changes, for example disturbance or reduced shading [1]; competition is only one possible cause, so the result shows association, not proven cause [1]. [5] Examiner insight: Controls and replication are what make an experiment convincing; mention them.

10. (a) The two snail species are distributed independently of each other [1] (b) Row total 25, column total 23: 25 × 23 ÷ 60 [1] = 9.58 [1] (c) (4 − 9.58)² ÷ 9.58 = 3.25; (21 − 15.42)² ÷ 15.42 = 2.02; (19 − 13.42)² ÷ 13.42 = 2.32; (16 − 21.58)² ÷ 21.58 = 1.44 [1]; χ² = 9.04 [1] (d) df = (2 − 1)(2 − 1) = 1 [1]; 9.04 exceeds 3.84, so reject the null: distribution is not independent (negative association) [1] (e) Both together is fewer than expected, suggesting competition [1], but the snails may prefer different conditions, so cause is not shown [1]. [9] Examiner insight: Show the expected-value method and write the conclusion in context, not only “reject”.

11. (a) Both populations cycle [1]; lynx peaks follow hare peaks by about a year [1] (b) More hares support more lynx [1]; more lynx reduce hares, then lynx fall from lack of food [1] (c) Top-down: predators limit prey; bottom-up: food supply limits populations [1]; hare food supply may also matter [1]. [6] Examiner insight: Use the words “delayed” and “negative feedback”, and describe the lag.

12. (a) Both release chemicals that deter competitors [1] (b) Allelopathy: black walnut releases juglone [1]. Antibiotic: Penicillium secretes penicillin [1] (c) Plants; microorganisms (fungi or bacteria) [1]. [4] Examiner insight: A named organism plus a named chemical is stronger than a general statement.

Where marks are usually lost

• Using the Lincoln index without clear M, N and R.
• Forgetting to scale quadrat density up to the whole area.
• Listing assumptions without the effect of breaking them.
• Missing which axis is logarithmic.
• Describing a plateau without balanced births and deaths.
• Concluding that a chi-squared result proves competition.
• Giving one-sided benefits for mutualism.
• Writing “reject” without saying what it means for the species.

Next steps

• Review the revision notes and study guide.
• Course hub: IB DP Biology; checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization. Section C4.1, Populations and communities.