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IB Biology B4.2: Ecological niches worked solutions

Worked solutions to every practice question in B4.2: Ecological niches 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 defines a heterotroph? (A) Makes its own organic molecules (B) Takes in organic molecules made by others (C) Uses only light as energy (D) Feeds only on dead matter

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B. A heterotroph takes in organic molecules that other organisms made. A describes an autotroph, C confuses the energy source with the carbon source, and D describes only saprotrophs, which are one kind of heterotroph.

Final answer

Question A2

[1 mark]

A fungus grows on a fallen log. Which description of its nutrition is correct? (A) Holozoic: ingests wood, digests inside (B) Detritivore: swallows dead matter (C) Saprotrophic: digests wood outside (D) Autotrophic: makes organic molecules

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C. A fungus on dead wood secretes enzymes onto the log, digests it outside the body and absorbs the products: saprotrophic nutrition. A and B both involve ingestion and internal digestion, and D is wrong because the fungus cannot make its own organic molecules.

Final answer

Question A3

[1 mark]

In a tube of jelly with oxygen entering from the top, a bacterium forms colonies only in the bottom 2 cm. Which is it? (A) Obligate aerobe (B) Facultative anaerobe (C) Mixotroph (D) Obligate anaerobe

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D. Colonies only where oxygen is scarce show that the bacterium grows without oxygen and is harmed by it: an obligate anaerobe. An obligate aerobe (A) would sit at the top, a facultative anaerobe (B) would grow throughout, and mixotroph (C) is a nutrition term, not an oxygen group.

Final answer

Question A4

[1 mark]

Which organism has mixotrophic nutrition? (A) A protist that photosynthesises and ingests bacteria (B) A fungus on dead wood (C) A moss on a wall (D) A lion hunting zebra

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A. A protist that photosynthesises and also ingests bacteria uses both an autotrophic and a heterotrophic mode. B is a saprotroph, C an autotroph and D a holozoic heterotroph.

Final answer

Question A5

[1 mark]

Which statement describes the realized niche of a species? (A) All conditions it could tolerate (B) Its role with no competitor (C) The part of its fundamental niche it uses (D) Where it was first found

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C. The realized niche is the part of the fundamental niche a species actually occupies when competitors are present. A describes the fundamental niche, B ignores competition and D is a habitat, not a niche.

Final answer

Question A6

[1 mark]

Two species with identical niches share one habitat for many generations. What does the competitive exclusion principle predict? (A) Both increase in number (B) Both stay constant (C) They interbreed (D) One declines and is eliminated

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D. If two species need exactly the same limited resource, the better competitor gains it and the other declines until it is eliminated from that habitat. Stable coexistence (A and B) requires different niches, and interbreeding (C) is not part of the principle.

Final answer

Question A7

[1 mark]

What is the main role of saprotrophs in an ecosystem? (A) Trapping light energy (B) Breaking down dead matter and recycling nutrients (C) Pollinating flowers (D) Fixing nitrogen in root nodules

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B. Saprotrophs digest dead organic matter and release mineral ions and carbon dioxide for reuse. They do not trap light (A), pollinate (C) or fix nitrogen in root nodules (D).

Final answer

Question A8

[1 mark]

An oak tree is eaten by beetle larvae, and a fungus grows on the fallen oak leaves. Which row gives the modes of nutrition? (A) oak autotroph; larva holozoic; fungus saprotroph (B) oak autotroph; larva saprotroph; fungus holozoic (C) oak mixotroph; larva holozoic; fungus saprotroph (D) oak heterotroph; larva holozoic; fungus autotroph

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A. The oak makes its own food by photosynthesis, the larvae ingest and digest leaves inside the body (holozoic), and the fungus digests dead leaves outside its body (saprotrophic). B swaps the last two, C calls the oak mixotrophic and D calls the fungus autotrophic.

Final answer

Question A9

[1 mark]

Which feature is an adaptation of prey for avoiding predation? (A) Forward-facing eyes (B) Long sharp claws (C) Fur matching the background (D) Hunting in a pack

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C. Fur that matches the background is camouflage, which makes the prey harder to see. A, B and D help a predator to find, hold or catch food.

Final answer

Tutorial practice

Question B1

[3 marks]

The climograph data are for one place. Use them to answer the questions. Monthly rain in mm from January to December: 220, 200, 250, 230, 210, 140, 120, 130, 170, 210, 240, 260. Monthly temperature in degrees C: 26, 26, 27, 28, 27, 26, 26, 26, 26, 26, 26, 26. (a) State the maximum precipitation. [1] (b) State the warmest month. [1] (c) Identify the biome. [1]

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(a) The highest precipitation in the table is 260 mm, in December (1). (b) The highest temperature is 28 ^{\circ}C, in April (1). (c) Temperature stays between 26 and 28 ^{\circ}C and every month has at least 120 mm of rain: warm and wet all year is tropical rainforest (1).

Final answer

(a) 260 mm, (b) April, (c) tropical rainforest.

Question B2

[5 marks]

A little egret feeds on small fish and crabs in a shallow estuary, mostly at low tide, and nests in tall trees nearby. (a) Outline what is meant by the ecological niche of the egret. [2] (b) State two biotic factors and one abiotic factor that influence its niche. [3]

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(a) The niche is the role of a species in its ecosystem, including where it lives, what it eats and its interactions with other species (1); a specific feature from the egret, such as feeding on fish and crabs in the shallow estuary at low tide (1). (b) Any two biotic factors: competition with other waders for the same fish (1), the abundance of its prey (1), predation or parasitism of the egret (1; maximum 2). One abiotic factor: tide level, salinity, temperature or water depth (1).

Final answer

(a) The niche is the role of the species in its ecosystem: where it lives and feeds, what it eats, when it is active and how it interacts with other species; for the egret, shallow estuary feeding on fish and crabs at low tide. (b) Biotic: competition from other wading birds for fish; the numbers of its prey fish. Abiotic: tide level or salinity.

Question B3

[4 marks]

A bracket fungus and an earthworm both obtain energy from dead leaves. Compare how the two organisms obtain their energy.

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Similarity: both are heterotrophs that obtain organic molecules from dead plant material (1). Fungus: secretes digestive enzymes onto the leaf, so digestion is external, and it then absorbs the small molecules: saprotrophic nutrition (1). Earthworm: ingests the dead leaf and digests it inside the gut with enzymes (1); the products are absorbed and assimilated: detritivore, a form of holozoic nutrition (1).

Final answer

Both are heterotrophs that take organic molecules from dead leaves. The fungus secretes enzymes onto the leaf, digests it outside the body and absorbs the products (saprotroph). The earthworm swallows leaf fragments and digests them inside its gut (detritivore, holozoic) before absorbing and assimilating the products.

Question B4

[5 marks]

Outline adaptations of herbivores for feeding on plants and adaptations of plants for resisting herbivory. A maximum of three marks may be given for either side.

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Herbivore side, any of (maximum 3): broad flat molars or chewing or piercing mouthparts to break plant tissue (1); gut microbes or enzymes that digest cellulose (1); detoxifying enzymes for plant toxins (1); a specific example such as the monarch caterpillar storing milkweed toxins (1). Plant side, any of (maximum 3): thorns or spines as a physical barrier (1); hard, rigid leaves and stems that are hard to chew (1); toxic secondary metabolites in leaves or seeds (1); camouflage (1); chemicals that attract the herbivore's predators (1). Total marks limited to 5.

Final answer

Herbivores: grinding teeth or chewing or piercing mouthparts; gut microbes or enzymes to digest cellulose; chemistry that detoxifies plant toxins. Plants: thorns or spines; hard, tough leaves and stems; toxic secondary metabolites; camouflage; attracting the herbivore's predators.

Question B5

[4 marks]

A tropical forest has tall canopy trees, lianas, epiphytes and small herbs on the dark forest floor. (a) Outline two different ways in which forest plants obtain enough light. [2] (b) Explain why an epiphyte is not a parasite of the tree it grows on. [2]

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(a) Two different strategies, each with its link to light (1 each): a liana roots in the soil and climbs a tree for support, so its leaves reach the canopy without building a thick trunk; an epiphyte grows on a high branch where light is better; a shade-tolerant herb has broad thin leaves and plenty of pigment to absorb the dim light. (b) The epiphyte uses the host tree only for physical support (1) and obtains water and minerals from rain, air and debris, not from the host's sap, so it does not harm the tree (1).

Final answer

(a) A liana roots in the soil and climbs a tree, so its leaves reach the sunlit canopy without a thick trunk. An epiphyte grows high on a branch, so it sits in better light. A forest-floor herb has broad thin leaves with much pigment to absorb dim light. (b) An epiphyte uses the host only for support and takes water and minerals from rain, air and debris, so it does not feed on the host.

Question B6

[5 marks]

A freshwater protist was grown for four days from 100 cells per cm^3 under four conditions. Cells per cubic centimetre after four days: light without prey 520, light with prey 900, dark with prey 380, dark without prey 40. (a) Deduce whether the protist is an obligate or a facultative mixotroph, giving evidence. [2] (b) Calculate the percentage change in dark conditions without prey. [1] (c) Explain why growth is greatest in light with prey. [2]

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(a) Facultative mixotroph (1): it grows with photosynthesis alone (520 against the start of 100) and with feeding alone (380), so either route is enough (1). (b) (40-100)/100\times100=-60\,\% (1). (c) In light with prey the protist both photosynthesises and feeds (1), so it has two sources of organic molecules and energy and grows fastest, 900 cells per cm^3 (1).

Final answer

(a) Facultative mixotroph: it grows with light alone and with food alone. (b) -60\,\%. (c) Light and food give two sources of carbon compounds and energy, so growth is greatest.

Question B7

[4 marks]

Three archaea live in different habitats. Archaeon X in a salt lake absorbs light with a membrane pigment and uses it to make ATP. Archaeon Y at a hot vent oxidises hydrogen gas to release energy and makes organic molecules from carbon dioxide. Archaeon W in sediment takes in organic compounds from dead cells and oxidises them to release energy. (a) Identify the source of energy for each archaeon. [3] (b) Explain why archaea cannot be described by a single mode of nutrition. [1]

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(a) X uses light (1); Y oxidises an inorganic chemical, hydrogen (1); W oxidises organic carbon compounds (1). (b) The three obtain energy for ATP in different ways, so archaea have a diversity of nutrition and cannot be given one mode (1).

Final answer

(a) X: light. Y: oxidation of an inorganic chemical (hydrogen gas). W: oxidation of carbon compounds. (b) Archaea show diversity of nutrition, so no single mode describes them as a group.

Question B8

[6 marks]

Three fossil mammal skulls, R, S and T, are compared. Skull R: small canines, very large flat ridged cheek teeth, thick enamel. Skull S: large pointed canines, narrow blade-like cheek teeth, thin enamel. Skull T: moderate canines, medium cheek teeth with rounded cusps, medium enamel. (a) Deduce the likely diet of each skull, with the tooth evidence. [3] (b) Suggest two other kinds of evidence that would support the deduction for R. [2] (c) State one limit of inferring diet from teeth alone. [1]

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(a) R has very large molars with thick enamel and small front teeth, which grind hard plant food: herbivore (1). S has large pointed canines and narrow blade-like cheek teeth, which pierce and shear flesh: carnivore (1). T has all four tooth types, none extreme, with rounded cusps: omnivore (1). (b) Any two: strontium-to-calcium ratio in the bones (1); microscopic pits and scratches on the enamel (1); butchered bones or plant remains found with the fossil (1); maximum 2. (c) The skull shows what the teeth were suited to, not exactly what the animal ate; diet may have varied or changed (1).

Final answer

(a) R: plant eater, grinding. S: meat eater, tearing and shearing. T: omnivore. (b) Two of: strontium-to-calcium ratio in bone; wear pits and scratches on the enamel; butchered bones or plant remains found with the fossil. (c) Teeth show what the animal was built to eat and what it chewed recently; they do not prove what it ate over its whole life.

Question B9

[5 marks]

Three warbler species were watched feeding in the same spruce wood. The table shows the percentage of each species' feeding records in each height band (100 records per species). Percentage of feeding records in the upper, middle and lower height bands: A 72, 24, 4; B 11, 76, 13; C 3, 21, 76. (a) State the height band where species C feeds most. [1] (b) Compare the use of height by species A and species C. [2] (c) Explain how the differences allow the three species to coexist. [2]

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(a) Species C, in the lower band (1). (b) A feeds mostly in the upper band, 72% of its records (1), whereas C feeds mostly in the lower band, 76%; only 3% of C's records are in the upper band, so their use of height barely overlaps (1). (c) Each species uses a different part of the tree, so competition for the same food is reduced (1) and, because their niches differ, competitive exclusion does not occur and all three coexist (1).

Final answer

(a) C, lower band. (b) A feeds mostly in the upper band (72%) and C in the lower band (76%); B feeds mostly in the middle band (76%), so overlap between A and C is small (3% against 4% of records). (c) They use different parts of the tree, so they compete less for food; each realized niche is narrower than its fundamental one.

Question B10

[5 marks]

Two species of flour beetle, P and Q, were kept separately and together in jars of the same flour. The table gives the number of adults after 40 days. Adults after 40 days. At 25 degrees C: P alone 210, Q alone 140, together P 190 and Q 0. At 30 degrees C: P alone 120, Q alone 230, together P 0 and Q 200. (a) State which species survives when the beetles share a jar at 25 ^{\circ}C. [1] (b) Explain this outcome. [2] (c) Suggest why the outcome is different at 30 ^{\circ}C. [2]

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(a) Species P survives at 25 ^{\circ}C (1). (b) Both species feed on the same flour, so they have the same niche in the jar (1); the competitive exclusion principle says one will be eliminated, and the better competitor at that temperature, P, remains (1). (c) Temperature changes how well each species grows and competes (1), so at 30 ^{\circ}C Q is the better competitor and excludes P (1).

Final answer

(a) P at 25 ^{\circ}C. (b) P and Q need the same food, so with an overlapping niche the better competitor wins and the other is eliminated. (c) At 30 ^{\circ}C the competitive balance changes, so Q wins.

Advanced practice

Question C1

[7 marks]

Two minnow species, X and Y, were placed in a tank with three depth zones, first each alone and then both together. The table gives the percentage of each species found in each zone. Percentage of fish in depth zones 0 to 20, 20 to 40 and 40 to 60 cm. Alone: X 30, 45, 25; Y 10, 35, 55. Together: X 60, 38, 2; Y 2, 8, 90. (a) State the zone where X is most abundant when alone. [1] (b) Calculate the change in the percentage of X in the 0 to 20 cm zone when Y is present. [1] (c) Use the data for X to distinguish its fundamental niche from its realized niche. [2] (d) Explain why X and Y coexist in the tank. [3]

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(a) 20 to 40 cm, 45% (1). (b) 60-30=+30 percentage points (1). (c) Alone, X uses all three zones, which shows its fundamental niche in depth (1); together it is almost confined to the shallow zones (98% in the top two), its realized niche, because Y takes the deeper water (1). (d) The two niches overlap, so if both used the same zone one would be eliminated (1); instead each keeps to the zone where it is the better competitor, X shallow and Y deep (1), so their realized niches differ and they coexist (1).

Final answer

(a) 20 to 40 cm. (b) +30 percentage points. (c) The alone distribution is the fundamental niche (all depths used); together each is limited to the part it keeps against the other: X to shallow water, Y to deep water. (d) The niches overlap in the middle zone, but each species keeps to the zone where it is the better competitor; with different realized niches, competitive exclusion does not occur.

Question C2

[7 marks]

Two protists, P and Q, were grown from 100 cells per cm^3 for four days. The table gives the final cell density. Cells per cubic centimetre after four days in light with prey, light only, dark with prey, dark only. P: 640, 410, 360, 30. Q: 700, 60, 50, 20. (a) Deduce whether each protist is an obligate or a facultative mixotroph, using data. [4] (b) Calculate the percentage change in P in light without prey. [1] (c) Suggest an advantage of facultative mixotrophy in a pond that is shaded in winter. [2]

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(a) P: facultative mixotroph (1), since it grows with light alone, 410, and with prey alone, 360 (1). Q: obligate mixotroph (1), since only light with prey gives growth, 700, and either route alone gives a fall to 60 or 50 cells per cm^3 (1). (b) (410-100)/100\times100=+310\,\% (1). (c) A facultative mixotroph can switch between photosynthesis and feeding (1), so in a pond that is shaded in winter it can feed when light is low and photosynthesise when prey is scarce (1).

Final answer

(a) P is a facultative mixotroph: it grows with light alone (410) or with prey alone (360). Q is an obligate mixotroph: it grows well only with light and prey together (700), and not with either alone. (b) +310\,\%. (c) In a pond that is dark in winter, P can feed on bacteria when light is low and photosynthesise when prey is scarce, so it can keep growing.

Question C3

[7 marks]

Plant S makes a bitter toxin. Beetle B eats plant S, stores the toxin and has bright red and black wing cases. Lizards hunt by sight and ate 64% of beetles fed on a non-toxic plant and 8% of beetles fed on plant S. (a) Calculate by how many percentage points the toxin reduced the share of beetles eaten. [1] (b) Explain how the toxin helps plant S. [1] (c) Explain how eating plant S protects the beetle. [2] (d) Suggest the role of the beetle's bright colour. [1] (e) Outline the niche of beetle B, using the passage. [2]

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(a) 64-8=56 percentage points lower (1). (b) The toxin is a chemical defence that makes the leaves bitter or poisonous, so herbivores eat less of the plant (1). (c) The beetle stores the toxin in its body (1), so a lizard that eats it is harmed or tastes it and learns to avoid similar beetles, which lowers the percentage eaten (1). (d) The bright colour is a warning signal to predators that the beetle is toxic (1). (e) The niche is the role of B in the ecosystem (1): it feeds on plant S and is the prey of lizards, avoiding them by toxin and warning colour (1).

Final answer

(a) 56 percentage points. (b) The bitter toxin makes the plant unpalatable, so fewer leaves are eaten. (c) The beetle stores the toxin, which makes it harmful or distasteful, so lizards that eat one learn to avoid it. (d) Bright colour is a warning signal that advertises the toxin. (e) The niche of B: it lives on and feeds on plant S, is active by day and has predators such as lizards that learn to avoid it.

Question C4

[8 marks]

Grass species G1 and G2 were grown alone and mixed in pots of dry soil and wet soil. The table gives the yield of each species in grams per pot. Yield in g per pot. Dry soil: G1 alone 40, G2 alone 25, G1 mixed 38, G2 mixed 5. Wet soil: G1 alone 55, G2 alone 60, G1 mixed 12, G2 mixed 50. (a) Describe how soil moisture limits G2 when it is grown alone. [1] (b) Calculate the percentage change in the yield of G1 when mixed with G2 in dry soil and in wet soil. [2] (c) Deduce the realized niche of each species in terms of soil moisture. [2] (d) Explain why both species persist in a field with both wet and dry patches. [2] (e) Predict the outcome in a field with only wet soil. [1]

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(a) G2 gives 60 g in wet soil but only 25 g in dry soil, so its fundamental niche is wetter soil than for G1 (1). (b) G1 in dry soil: (38-40)/40\times100=-5\,\% (1). G1 in wet soil: (12-55)/55\times100=-78\,\% (1). (c) G1 is hardly reduced by G2 in dry soil (38 against 40) so its realized niche is dry soil (1); G2 yields 50 g in wet soil but only 5 g in dry soil when mixed, so its realized niche is wet soil (1). (d) Each species is the better competitor in one soil type (1); in a field with both, each persists in its own patches, so exclusion does not remove either (1). (e) In all-wet soil G2 would be the better competitor and G1 would be excluded (1).

Final answer

(a) G2 is limited by dryness: 25 g in dry soil against 60 g in wet. (b) -5\,\% in dry soil and -78\,\% in wet soil. (c) G1 has the realized niche in dry soil, where it is hardly reduced; G2 has the realized niche in wet soil. (d) In a field with wet and dry patches each species wins in its own patch type, so competitive exclusion does not remove either. (e) In an all-wet field G2 would exclude G1.

Connected questions

Question X1

Connected questions[6 marks]

B4.2 + C4.2 + C1.3. In an oak wood, energy passes from oak leaves to caterpillars to blue tits. Fallen leaves are broken down by fungi, and a shade-tolerant herb grows beneath the oak. (a) Explain why less energy is available to the blue tits than to the caterpillars. [2] (b) Explain how the fungus helps the oak to photosynthesise and grow. [2] (c) Suggest two features of the herb's leaves that help it to harvest light. [2]

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(a) Energy is lost as heat from respiration (1) and in parts that are uneaten or egested, so the energy stored at each trophic level is smaller (1). (b) The fungus respires and releases carbon dioxide, which the oak uses in photosynthesis (1); its enzymes digest leaf litter and release mineral ions that the oak's roots absorb (1). (c) Broad thin leaves give a large area to absorb the little light (1), and extra chlorophyll captures more of it, so the herb photosynthesises enough to survive in shade (1).

Final answer

(a) Energy is lost at each step as heat from respiration, in uneaten parts and in faeces, so less is stored in each higher level. (b) The fungus digests leaf litter and releases carbon dioxide by respiration, which the oak uses in photosynthesis, and minerals that the roots absorb. (c) Broad thin leaves with much chlorophyll absorb more of the dim light, so more photosynthesis occurs.

Question X2

Connected questions[6 marks]

B4.2 + C4.1 + D4.1. Two seed-eating finch species live on separate islands, with the same beak depth of about 9 mm. On a third island where both live together, one species has a beak 7 mm deep and the other 11 mm. (a) Explain how competition could have caused the difference between the beaks. [3] (b) Explain why the two species can now coexist. [1] (c) Explain how intraspecific competition limits the size of each finch population. [2]

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(a) There is variation in beak depth (1); birds whose beaks differ from the other species' beaks find more seeds with less competition, so they survive and reproduce more (1); the alleles for those beaks increase in later generations, so beak depths diverge (1). (b) The two species eat different sizes of seed, so their realized niches differ and competitive exclusion does not occur (1). (c) Members of the same species compete for the same limited seeds (1), so as density rises the population grows more slowly and levels off at the carrying capacity (1). Mark total limited to 6.

Final answer

(a) Birds whose beaks differ from the other species' beaks eat seeds that face less competition, so they survive and reproduce more and pass the alleles on; beak depths diverge. (b) Both species have different beaks, so they eat seeds of different sizes: their niches differ and exclusion is avoided. (c) Within a species, individuals compete for the same seeds, so population growth slows as density rises and stops at the carrying capacity.

Question X3

Connected questions[6 marks]

B4.2 + C1.2 + A2.2. A bacterium from a soil sample grows well in air but not at all in a sealed jar without oxygen. Yeast cells grow in both conditions but multiply faster in air. (a) Identify the oxygen group of the soil bacterium and explain why it cannot grow without oxygen. [2] (b) Explain why yeast multiplies faster in air. [2] (c) State where the bacterium carries out aerobic respiration, given that it has no mitochondria. [2]

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(a) Obligate aerobe (1); it needs oxygen as the final electron acceptor in respiration to produce ATP, and it does not ferment (1). (b) Aerobic respiration yields far more ATP per glucose than fermentation (1), so more energy is available for growth and division in air (1). (c) Bacteria are prokaryotes and have no mitochondria (1); the electron transport chain is in the plasma membrane (1).

Final answer

(a) The bacterium is an obligate aerobe: it needs oxygen as the final electron acceptor to make ATP. (b) With oxygen yeast respires aerobically and makes much more ATP per glucose than anaerobic fermentation, so it grows faster. (c) A bacterium has no mitochondria; its electron transport chain is in the plasma membrane.

From the book

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

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

About the book