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IB Biology C4.1: Populations and communities worked solutions

Worked solutions to every practice question in C4.1: Populations and communities of Photon Biology: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

[1 mark]
Which statement describes a community? (A) All the members of one species in a habitat (B) All the interacting populations in an area (C) All the living and non-living things in an area (D) All the members of a species worldwide
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A community is all the populations of different species that interact in an area. Option A is a population, C an ecosystem.

Final answer

B

Question A2

[1 mark]
Two grasses in one lawn each grow less than when alone. Which relationship is this? (A) Mutualism (B) Herbivory (C) Interspecific competition (D) Parasitism
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Both species are harmed because they use the same limited resources, so this is interspecific competition.

Final answer

C

Question A3

[1 mark]
Which factor limiting a population is density-dependent? (A) A late frost (B) Spread of disease in a crowded herd (C) A volcanic eruption (D) A flood
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Disease spreads more easily when individuals are crowded, so its effect rises with density. The others act whatever the density.

Final answer

B

Question A4

[1 mark]
A population has levelled off at its carrying capacity. Which statement is correct? (A) births + immigration > deaths + emigration (B) births + immigration = deaths + emigration (C) deaths exceed births (D) there is no immigration
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At the plateau the size is steady, so the gains from births and immigration equal the losses from deaths and emigration.

Final answer

B

Question A5

[1 mark]
In a mark and recapture study M=80, N=60 and R=12. What is the Lincoln index estimate? (A) 16 (B) 128 (C) 400 (D) 1600
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\frac{M\times N}{R}=\frac{80\times60}{12}=400.

Final answer

C

Question A6

[3 marks]
Five random quadrats, each 0.5\,\mathrm{m^2}, were placed on a lawn of area 200\,\mathrm{m^2}. They held 2, 5, 3, 4 and 6 daisies. (a) Calculate the mean number of daisies per quadrat. [1] (b) Estimate the number of daisies on the lawn. [2]
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(a) 20\div5=4 per quadrat (1). (b) The lawn holds 200\div0.5=400 quadrat areas (1), so 4\times400=1600 daisies (1).

Final answer

(a) 4. (b) 1600 daisies.

Question A7

[1 mark]
Plots X and Y both have a mean of 6 plants per quadrat. The SD for X is 0.8 and for Y is 4.1. What does this show? (A) Y holds more plants (B) X is more evenly spread than Y (C) X is patchier than Y (D) The patterns are the same
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A small SD means the counts are similar from quadrat to quadrat, so the plants are evenly spread. Y is patchy.

Final answer

B

Question A8

[1 mark]
Presence or absence of two species is recorded in 50 quadrats. Which test checks for association? (A) t-test (B) chi-squared test (C) Standard deviation (D) Mean count per quadrat
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Presence and absence give counts in categories, and the chi-squared test compares observed with expected counts.

Final answer

B

Question A9

[1 mark]
Bacterial numbers on a log vertical axis against time form a straight rising line. What does this show? (A) Growth is exponential (B) The population has reached carrying capacity (C) Growth is slowing (D) The population is declining
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Equal ratios give equal steps on a log axis, so a straight rising line means exponential growth.

Final answer

A

Tutorial practice

Question B1

[4 marks]
Hard corals contain photosynthetic algae called zooxanthellae. (a) State what each partner gains from the relationship. [2] (b) Explain what happens to a coral that loses its algae in unusually warm water. [2]
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(a) The coral gains sugars (organic compounds) made by the algae in photosynthesis (1); the algae gain shelter and a place in the light, with nutrients from the coral (1). (b) Without the algae the coral loses much of its food supply (1), so it has less energy to grow and reproduce and may die (1).

Final answer

(a) Coral gets sugars from the algal photosynthesis; algae get shelter and light in the coral. (b) The coral loses its main food source, so it grows and reproduces less and may die.

Question B2

[4 marks]
Predation and parasitism are interspecific relationships. (a) Distinguish between predation and parasitism. [2] (b) Give a named example of each. [2]
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(a) A predator kills the prey and eats it (1); a parasite lives on or in a host, feeding from it and harming it but usually keeping it alive for some time (1). (b) Predation: for example barn owl and field vole (1); parasitism: for example tapeworm in a dog or a tick on a mammal (1). Disease-causing bacteria are not accepted as the parasite example.

Final answer

(a) A predator kills and eats its prey; a parasite lives on or in a host and feeds from it without usually killing it quickly. (b) Owl and vole; tapeworm in the gut.

Question B3

[4 marks]
Hundreds of seedlings of one tree species germinate under the parent tree. (a) Explain why competition between these seedlings is intense. [2] (b) Outline one example of cooperation between members of a single species. [2]
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(a) The seedlings need the same resources, such as light, water and mineral ions (1), and these are in limited supply where so many grow close together (1). (b) Name the species and the task, for example meerkats where a sentinel gives an alarm call (1); the others escape, so each member gains a better chance of survival (1).

Final answer

(a) They need identical resources, light, water, minerals, and these are limited at high density. (b) Meerkat sentinel gives alarm calls so the group escapes.

Question B4

[5 marks]
A student estimates the number of woodlice in a garden using capture, mark, release and recapture. (a) Outline the method. [3] (b) State two assumptions of the method. [2]
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(a) Any three. Catch a random sample, count it and mark each woodlouse harmlessly (1); release them and leave time to mix with the others (1); take a second random sample and count all of it and the marked ones (1); calculate M\times N/R (1). (b) Any two. No immigration or emigration (1); few births and deaths between the samples (1); marks are permanent and do not affect survival (1); marked animals mix randomly (1).

Final answer

(a) Catch a random sample and count; mark harmlessly and release; wait for mixing; second sample, count total and marked; use M\times N/R. (b) No migration; marks not lost or harmful.

Question B5

[3 marks]
Ecologists estimate a seabird colony by counting occupied nests. (a) Suggest one reason why the number of nests does not give the number of birds exactly. [1] (b) State one advantage of counting nests rather than catching and marking the birds. [1] (c) Suggest how the sampling error could be reduced. [1]
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(a) Non-breeding birds have no nest, and each nest holds more than one bird (1). (b) Birds are not caught, so there is little disturbance, and a count is quick (1). (c) Repeat the count and sample many areas chosen at random (1).

Final answer

(a) Non-breeding birds have no nest; a nest holds a pair and chicks. (b) Less disturbance, quick. (c) Count several times or sample many areas at random.

Question B6

[5 marks]
A few yeast cells are added to a sugar solution in a closed flask and counted over several days. (a) Describe the curve of cell number against time, naming each phase. [3] (b) Explain why the number of cells stops rising. [2]
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(a) Cell number rises with ever steeper increase: exponential growth (1); the rate of increase then slows (1); finally it levels at a plateau near carrying capacity (1). (b) Resources such as sugar run short or waste such as ethanol builds up (1), so deaths rise and births fall until they are equal (1).

Final answer

(a) Rises slowly then ever faster (exponential), growth slows, then plateau. (b) Sugar runs short and waste builds up, so deaths equal births.

Question B7

[6 marks]
Duckweed was grown in tanks of the same nutrient solution in full light or in shade. Fronds were counted at intervals. Day 0, 7, 14, 21, 28. Fronds in full light: 20, 55, 140, 250, 285. Fronds in shade: 20, 36, 66, 100, 112. (a) Compare the two growth curves. [2] (b) Calculate the percentage increase in the number of fronds in full light between day 7 and day 14. [2] (c) Suggest two factors that could limit the full-light population on day 28. [2]
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(a) Both rise and level off (1); full light is higher at every time and levels at a higher number (1). (b) \frac{140-55}{55}\times100 (1) =154.5\approx155\% (1). (c) Any two: space (1); nutrients, shading by other fronds, carbon dioxide or pH (1).

Final answer

(a) Both sigmoid; full light higher at every time, plateau higher. (b) 155%. (c) Space; nutrients (also light, carbon dioxide).

Question B8

[5 marks]
In 80 random quadrats the presence of species X and species Y was recorded. Both were present in 22; only X in 18; only Y in 18; neither in 22. X was present in 40 quadrats and Y in 40. (a) State the null hypothesis. [1] (b) Calculate the expected number of quadrats containing both species if they are independent. [1] (c) The four expected values are all equal. Calculate \chi^2. [2] (d) The critical value for 1 degree of freedom at p=0.05 is 3.84. State the conclusion. [1]
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(a) The presence of X is independent of the presence of Y (1). (b) \frac{40\times40}{80}=20 (1). (c) E=20 in each cell; \chi^2=\frac{(22-20)^2}{20}+\frac{(18-20)^2}{20}+\frac{(18-20)^2}{20}+\frac{(22-20)^2}{20}=0.2+0.2+0.2+0.2 (1) =0.8 (1). (d) 0.8<3.84, so there is no significant association; keep the null hypothesis (1).

Final answer

(a) X and Y occur independently. (b) 20. (c) 0.8. (d) No significant association.

Question B9

[4 marks]
Barnacle species A lives on rock plots. B was removed from ten plots and left in ten control plots. After a year the mean density of A was 46\,\mathrm{m^{-2}} with B removed and 31\,\mathrm{m^{-2}} with B present. (a) State what the result indicates. [2] (b) Explain why it does not prove interspecific competition. [1] (c) Name one other approach to testing for competition. [1]
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(a) A reaches a higher density when B is absent (1), which indicates competition between A and B for a limited resource (1). (b) Other factors, such as disturbance by the removal or differences between plots, could cause the change (1). (c) A laboratory experiment growing the species alone and together, or random field sampling for association (1).

Final answer

(a) A does better without B, which indicates competition for a shared resource. (b) Other differences between plots or the removal itself may be the cause. (c) A laboratory experiment with the species alone and together.

Question B10

[4 marks]
Prey and predator numbers (hundreds) in an area were counted each year. Year 0 to 8. Prey: 40, 70, 110, 90, 50, 30, 55, 100, 95. Predators: 8, 10, 16, 26, 32, 22, 12, 14, 22. (a) State the year of each peak and the delay of the predator peak. [2] (b) Explain the fall in prey numbers between year 2 and year 5. [2]
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(a) Prey peak in year 2 and predators in year 4 (1); the predator peak lags 2 years behind (1). (b) Predator numbers rose after the prey peak (1), so more prey were eaten than were born, and prey numbers fell (1).

Final answer

(a) Prey peak year 2, predator peak year 4: the predator peak is 2 years later. (b) More predators eat more prey, and the prey population falls.

Question B11

[4 marks]
Foxes eat rabbits, and rabbits eat grass. All the foxes in a grassland are removed. (a) Predict the effect on the rabbits and on the grass, with a reason. [3] (b) State whether this is top-down or bottom-up control. [1]
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(a) Rabbits increase because they are no longer eaten by foxes (1); more rabbits graze the grass (1), so the grass decreases (1). (b) Top-down, because control is by consumers (1).

Final answer

(a) Rabbits increase (no predation); grass decreases (more grazing). (b) Top-down.

Advanced practice

Question C1

[6 marks]
Water hyacinth was introduced to many tropical lakes. It forms thick floating mats. (a) Discuss how an introduced plant such as this can become invasive and affect the lake ecosystem. [6]
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Any six. It spreads quickly by vegetative reproduction (1) and has no natural herbivores or diseases in the new lake (1). It competes with native plants for light and nutrients (1); mats shade submerged plants and algae (1), so photosynthesis and oxygen fall (1). Fish and invertebrates die or leave (1), so biodiversity falls (1). A control such as a biological agent may reduce it, but the agent could itself harm native species (1).

Final answer

Fast vegetative spread, no natural herbivores, shades out natives, less light, less oxygen, fish and plant loss; control and its risk.

Question C2

[6 marks]
Bacteria in a flask were counted each hour. Hour 0 to 7. Count (thousands per cm3): 2.0, 4.1, 8.0, 16.2, 31.5, 52, 61, 63. (a) Show that growth was exponential in the first four hours. [2] (b) Describe how the first four points look on a graph with a logarithmic vertical axis. [2] (c) Explain the change in growth after hour 4. [2]
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(a) Count the ratio of successive counts: 4.1\div2=2.05, 8.0\div4.1=1.95, 16.2\div8.0=2.03 and 31.5\div16.2=1.94 (1); the count about doubles each hour, which is exponential (1). (b) The points lie close to a straight rising line (1), because equal ratios give equal steps on a log axis (1). (c) Food runs short and waste builds up (1), so deaths rise and births fall and the growth slows towards a plateau (1).

Final answer

(a) Each hour the count about doubles (2.05, 1.95, 2.03, 1.94). (b) A straight line. (c) Resources run short so growth slows.

Question C3

[6 marks]
Bluebells were counted in ten random 1\,\mathrm{m^2} quadrats in each of two plots. Plot P: 6, 5, 7, 6, 5, 6, 7, 5, 6, 7. Plot Q: 0, 14, 1, 12, 0, 13, 2, 11, 1, 6. For P the SD is 0.8; for Q the SD is 5.9. (a) Estimate the number of bluebells in the 500\,\mathrm{m^2} of plot P. [2] (b) Deduce what the SDs show about the two distributions. [2] (c) Suggest why the estimate for Q is less reliable and how to improve it. [2]
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(a) The mean for P is 60\div10=6 per m^2 (1); 6\times500=3000 (1). (b) P has a small SD, so the bluebells are spread evenly (1); Q has a large SD, so they are patchy (1). (c) The counts in Q vary widely, so the mean depends strongly on where the quadrats fall (1); use many more quadrats, placed at random (1).

Final answer

(a) mean 6; 3000. (b) P even; Q patchy. (c) Counts vary widely, so the sampling error is greater; use more quadrats.

Question C4

[6 marks]
Lettuce seeds were soaked in extract from the leaf litter of a tree and germination was recorded. Extract g per dm3: 0, 5, 10, 20, 40. Germination %: 92, 85, 61, 33, 8. (a) Describe the effect of the extract concentration on germination. [2] (b) Calculate the percentage decrease in germination at 20\,\mathrm{g\,dm^{-3}} compared with the control. [2] (c) Explain how the tree could benefit from releasing such a chemical. [2]
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(a) Germination falls as the concentration rises (1), from 92% with none to 8% at 40 g dm^{-3} (1). (b) \frac{92-33}{92}\times100 (1) =64.1\approx64\% (1). (c) The chemical (allelopathy) reduces germination of competing plants (1), so the tree and its seedlings face less competition for light, water and mineral ions (1).

Final answer

(a) Germination falls as concentration rises, from 92% to 8%. (b) 64%. (c) It kills seeds of competitors, so the tree has more light, water and minerals.

Question C5

[5 marks]
45 beetles were caught, marked with paint and released. A week later 60 beetles were caught and 9 had paint. (a) Calculate the Lincoln index estimate. [2] (b) Some of the paint wore off between the samples. Predict the effect on the estimate, with a reason. [2] (c) Explain why the estimate is not the true population size. [1]
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(a) \frac{45\times60}{9} (1) =300 beetles (1). (b) R is too small because some marked beetles are not recognised (1), so the estimate is too high (1). (c) It is based on samples, so it has a sampling error (1).

Final answer

(a) 300. (b) Too high, because fewer marked beetles are recognised so R is too small. (c) It comes from a sample, so there is sampling error.

Connected questions

Question X1

Connected questions[7 marks]
C4.1 + B4.2. Two freshwater snails, A and B, feed on the same algae. Alone, each reaches a stable number. Together, A stays stable and B falls to zero. (a) Explain what the result indicates about the niches of A and B. [2] (b) Explain why the realised niche of B differs from its fundamental niche. [2] (c) Outline a field removal experiment to test whether A causes the decline of B. [3]
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(a) The niches overlap, as both use the same food (1), so competition indicates competitive exclusion of B by A (1). (b) The fundamental niche is where B could live without competitors (1); the realised niche is smaller because A excludes it (1). (c) Remove A from some plots chosen at random (1) and leave control plots (1); count B in both at intervals and compare (1).

Final answer

(a) Overlapping niches; competitive exclusion; A out-competes B. (b) B can use the algae alone but A excludes it where they meet. (c) Remove A from random plots, keep control plots, compare B.

Question X2

Connected questions[7 marks]
C4.1 + C4.2. In a lake the energy entering each level is 12 000 kJ m^{-2} y^{-1} in algae, 1500 in zooplankton and 160 in perch. (a) Calculate the percentage of the energy in the algae that reaches the zooplankton. [1] (b) Explain why there are fewer perch than zooplankton. [2] (c) Perch are removed. Predict the effect on the zooplankton and the algae. [2] (d) Explain how extra phosphate would be a bottom-up effect. [2]
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(a) \frac{1500}{12\,000}\times100=12.5\% (1). (b) Energy is lost as heat in respiration and in waste at each level (1), so less energy is available to support numbers at higher levels (1). (c) Zooplankton increase because fewer are eaten (1), so more grazing makes the algae decrease (1). (d) Phosphate is a resource that limits algal growth (1); more algae supports more zooplankton and so on up the chain (1).

Final answer

(a) 12.5%. (b) Energy is lost at each step, so less supports perch. (c) Zooplankton increase, algae decrease. (d) More phosphate raises algal growth, which feeds the whole chain.

Question X3

Connected questions[7 marks]
C4.1 + D4.1 + C3.2. A soil mould secretes an antibiotic. Some bacteria near it carry an allele that makes them resistant. (a) Explain how secreting an antibiotic benefits the mould. [2] (b) Explain how natural selection could make most of the bacteria resistant. [3] (c) Explain why overuse of antibiotics in medicine favours resistant bacteria. [2]
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(a) The antibiotic kills or stops the growth of competing bacteria (1), so the mould has more nutrients and space (1). (b) The bacteria vary, with a few carrying the resistance allele (1); the resistant survive and reproduce more (1); the allele is passed on, so its frequency rises over generations (1). (c) Frequent use kills the non-resistant bacteria (1), leaving the resistant ones to multiply with little competition (1).

Final answer

(a) Kills competing bacteria, so the mould has more food and space. (b) Variation; resistant survive and reproduce, pass on allele. (c) It removes non-resistant bacteria, so resistant ones multiply.

Question X4

Connected questions[5 marks]
C4.1 + D4.3. Warming lowers the dissolved oxygen in a lake where cold-water char live. A survey marked 200 char; a second catch of 150 held 25 marked fish. (a) Calculate the Lincoln index estimate of the char population. [1] (b) Suggest why warming lowers the carrying capacity for char. [2] (c) Explain how density-dependent factors then change the char numbers. [2]
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(a) \frac{200\times150}{25}=1200 char (1). (b) Less oxygen is available for respiration (1), so the lake can support fewer fish (1). (c) If numbers exceed the new carrying capacity, competition for oxygen and food and disease intensify (1), so deaths rise and births fall until the numbers settle near the lower capacity (1).

Final answer

(a) 1200. (b) Less oxygen limits respiration, so fewer fish can survive; food may also fall. (c) Crowding and competition for oxygen increase mortality until numbers match the new lower K.

From the book

Photon Biology: Study Guide for the IB Diploma (SL & HL)

Populations and communities from the core ideas to full exam questions, with question types, worked examples and graded practice.

About the book

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