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IB Biology B4.1: Adaptation to environment worked solutions

Worked solutions to every practice question in B4.1: Adaptation to environment 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]

Researchers want to explain why a freshwater snail lives in some stretches of a river and not in others. Which is an abiotic factor that could limit where the snail lives? (A) Predatory fish numbers (B) Water temperature (C) Algae supply (D) Parasitic worm infection

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B. Water temperature is a non-living physical variable. A predatory fish (A), the algae the snail eats (C) and a parasitic worm (D) are all living things, so they are biotic factors.

Final answer

Question A2

[1 mark]

A desert beetle has a pale, waxy body surface, shelters under stones in the heat of the day and makes water when it digests fat. Which row classifies these three features in order? (A) Structural, behavioural, physiological (B) Behavioural, structural, physiological (C) Structural, physiological, behavioural (D) Physiological, behavioural, structural

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A. A waxy surface is a body part (structural), sheltering is something the beetle does (behavioural) and making water from fat is an internal chemical process (physiological). B, C and D give a different order of kinds.

Final answer

Question A3

[1 mark]

Which feature helps a red mangrove to live in waterlogged mud? (A) Aerial roots (B) A deep tap root (C) Rolled leaves (D) Large flat leaves

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A. Aerial roots stand above the mud and take in oxygen that the buried roots lack. Deep roots for fresh water (B), rolled leaves (C) and large flat leaves for dim light (D) do not solve the problem of oxygen-poor mud.

Final answer

Question A4

[1 mark]

Which pair of changes would both inhibit the formation of a coral reef? (A) Clearer water; a rise from 26 to 28 ^\circC (B) Muddy run-off; a fall to 14 ^\circC (C) Clearer water; a fall to 14 ^\circC (D) Muddy run-off; a rise from 26 to 28 ^\circC

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B. Muddy run-off cuts the light that reaches the symbiotic algae, and a fall to 14 ^\circC is below the 23 to 34 ^\circC that reef formation needs. In A the water is clearer and 28 ^\circC is within the range, so neither change inhibits a reef, and C and D each contain one change that does not.

Final answer

Question A5

[1 mark]

A weather station records a mean annual temperature of 27 ^\circC and 280 cm of precipitation a year. Which biome is most likely? (A) Desert (B) Tundra (C) Tropical rainforest (D) Taiga

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C. High temperature with very high precipitation is a tropical rainforest. A desert is hot but dry, the tundra is cold and dry, and the taiga is cold.

Final answer

Question A6

[1 mark]

A biome has long, very cold winters, a short summer and forests of conifers. Which biome is it? (A) Tundra (B) Taiga (C) Temperate forest (D) Grassland

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B. Long cold winters and conifers describe the taiga. The tundra has no trees, the temperate forest has milder winters and broad-leaved trees, and grassland has few trees.

Final answer

Question A7

[1 mark]

Which feature of a leaf would reduce water loss the most? (A) More stomata (B) A thin cuticle (C) Sunken stomata (D) A larger leaf area

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C. Stomata in sunken pits sit in a pocket of humid air, so less water vapour diffuses out. More stomata (A), a thin cuticle (B) and a larger surface area (D) all increase water loss.

Final answer

Question A8

[1 mark]

A water plant grows at pH 4 to pH 10 and not outside that range. Its growth is highest at pH 7. What is its range of tolerance for pH? (A) pH 4 to 10 (B) pH 7 only (C) pH 0 to 7 (D) pH 6 to 8

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A. The range of tolerance is the span between the two limits, pH 4 to pH 10. pH 7 is the optimum, not the range (B), and C and D give a span that does not match the limits.

Final answer

Question A9

[1 mark]

Which statement best describes the habitat of the sand lizard? (A) Hunts at dawn (B) Active in summer (C) Sandy heathland, England (D) Eaten by birds

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C. A habitat is the place where an organism lives, with its location, physical surroundings and ecosystem. A, B and D describe what the lizard does or what eats it, not where it lives.

Final answer

Question A10

[1 mark]

Four sites are surveyed as possible places for a coral reef. Which site is the best? (A) 5 m, 28 ^\circC, clear (B) 5 m, 28 ^\circC, cloudy (C) 50 m, 28 ^\circC, clear (D) 5 m, 15 ^\circC, clear

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A. It is shallow, warm and clear, so light reaches the algae and the temperature is within 23 to 34 ^\circC. B is cloudy, C is too deep for light and D is too cold.

Final answer

Tutorial practice

Question B1

[4 marks]

A desert shrub has small leaves, a thick waxy cuticle, stomata that open only at night and widely spreading shallow roots. (a) Explain how each of the four features helps the shrub to survive in the desert. [4]

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The small leaves have a small surface area, so less water evaporates (1); the waxy cuticle forms a waterproof barrier, so less water is lost through the leaf surface (1); the air is cooler at night, so evaporation from the open stomata is slower (1); wide, shallow roots cover a large area of soil and absorb water from brief rainfall before it drains away (1).

Final answer

Small leaves give a small surface area for evaporation; the waxy cuticle is a barrier to water loss; at night it is cool, so less water leaves each open stoma; wide shallow roots catch water from brief rain before it drains away.

Question B2

[4 marks]

A grass grows on lower mountain slopes but not above 2000 m, where it is colder, the soil is thin and the wind is stronger. (a) Explain how two of these abiotic factors could set the upper limit of the grass. [4]

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Two factors, each with an explanation (2 + 2). Lower temperature (1) slows enzyme-controlled reactions such as photosynthesis, so the grass cannot grow or reproduce (1). Stronger wind (1) increases transpiration and may damage or uproot the grass (1). Thin soil (1) holds less water and fewer mineral ions, so the roots cannot supply the plant (1). Accept any two of these, with the effect on the plant.

Final answer

Any two factors, each with its effect. Temperature: lower temperature slows enzyme activity and photosynthesis, so growth is too slow. Wind: raises transpiration and damages or uproots plants. Thin soil: holds little water and few mineral ions.

Question B3

[4 marks]

A transect ran inland from the sea over dunes. Mean counts of two plants per quadrat and soil salt content are shown. Distance from sea / m | 0 | 20 | 40 | 60 | 80; Species P / per quadrat, 0, 2, 7, 13, 15; Species Q / per quadrat, 14, 12, 6, 2, 0; Soil salt / g per kg, 12, 8, 4, 2, 1.5 (a) Describe the change in the abundance of species Q with distance from the sea. [2] (b) State the correlation between salt content and species P, and explain why it does not prove cause. [2]

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(a) Species Q decreases from 14 at 0 m (1) to 0 at 80 m, falling steadily (1). (b) Species P rises from 0 to 15 as the salt content falls from 12 to 1.5 g per kg, so there is a negative correlation (1); other variables such as soil water and wind also change along the transect, so the data do not prove that salt is the cause (1).

Final answer

(a) Species Q falls steadily from 14 per quadrat at 0 m to 0 at 80 m. (b) A negative correlation between salt and P; other factors such as wind and soil water also change inland, so salt is not proved as the cause.

Question B4

[5 marks]

Seedlings of a grass and a moss, each starting at 10 mg dry mass, grew for six weeks in full light or in 20 % light. Final dry masses in full light and in 20 % light: grass 52 mg and 21 mg; moss 24 mg and 27 mg. (a) Calculate the percentage change in dry mass of the grass in 20 % light. [1] (b) Describe the effect of shade on each species, and deduce which is commoner in woodland shade, with a reason. [4]

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(a) (21-10)/10\times100=110\,\% (1). (b) Shade reduced the dry mass of the grass from 52 mg to 21 mg (1), but the moss gained a little more mass in shade, 27 mg against 24 mg (1). The moss would be more abundant on a shaded woodland floor (1), because its growth in shade is as good as in full light (170 % against 140 %), while the grass grows much less (110 % against 420 %) (1).

Final answer

(a) 110 %. (b) Shade reduced the grass mass from 52 to 21 mg, but the moss mass rose slightly from 24 to 27 mg. Moss would be more abundant, since it grew better in shade (170 % against 140 %) while the grass grew much less in shade.

Question B5

[4 marks]

A desert tortoise lives mostly in a burrow and stores water in its bladder; the ground exceeds 50 ^\circC by day and rain is rare. (a) State the kind of adaptation shown by living in a burrow. [1] (b) Explain how the burrow and the water store help the tortoise to survive. [3]

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(a) Behavioural, because the tortoise is doing something (1). (b) The burrow is cooler than the surface, so the tortoise gains less heat (1), and its humid air means less water evaporates from the body (1). The stored water replaces what is lost during dry periods, so it can survive without drinking (1).

Final answer

(a) Behavioural. (b) The burrow is cooler and more humid than the surface, so less heat is gained and less water evaporates. The stored water replaces losses when there is no drinking water, so the tortoise survives dry spells.

Question B6

[4 marks]

A tall rainforest tree has buttress roots and drip-tip leaves; an orchid on its branch has roots that hang in the air. (a) Explain how buttress roots help a tall tree in the rainforest. [2] (b) Suggest one advantage of the drip tips and one of the orchid growing on a branch. [2]

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(a) The soil is thin, so buttress roots spread out near the surface (1) and support a tall trunk (1). (b) Water drains quickly from the drip tips, so fewer algae and fungi grow on the wet leaf surface (1). Growing on a branch lifts the orchid to the light, and the hanging roots take in rainwater and minerals without soil (1).

Final answer

(a) The soil is thin, so buttress roots spread out near the surface and support the tall trunk. (b) Drip tips let rain run off, so fewer algae and fungi grow on the leaves; the orchid is lifted to the light on a branch and its hanging roots take in rain water.

Question B7

[5 marks]

The growth rates of two fish species were measured at different water temperatures. Temperature / ^\circC | 5 | 10 | 15 | 20 | 25 | 30 | 35; Growth, species A / mm per week, 0, 1.5, 3.8, 5.0, 3.5, 0.8, 0; Growth, species B / mm per week, 0, 0, 1.0, 2.6, 4.4, 5.0, 0.5 (a) State the optimum temperature for species A and the range of temperatures over which it grew. [2] (b) Deduce, with a reason, which species is more likely to live in warmer water. [2] (c) Suggest why the growth of species A falls above its optimum. [1]

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(a) The optimum is 20 ^\circC, where growth is highest at 5.0 mm per week (1); species A grew at 10 to 30 ^\circC (1), and not at 5 or 35 ^\circC. (b) Species B (1), because its optimum is 30 ^\circC compared with 20 ^\circC and it still grows at 35 ^\circC when A does not (1). (c) Enzymes in the fish work less well or denature above the optimum, so growth falls (1).

Final answer

(a) Optimum 20 ^\circC; grew from 10 to 30 ^\circC. (b) Species B: optimum 30 ^\circC against 20 ^\circC, and still grows at 35 ^\circC. (c) Above the optimum enzymes begin to denature or work less well.

Question B8

[4 marks]

Station Q has a mean annual temperature of -8\,{}^\circC and 25 cm of precipitation a year; station R has 24 ^\circC and 15 cm. (a) Identify the biome at station Q and describe its vegetation. [2] (b) Identify the biome at station R and state one adaptation of its plants. [2]

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(a) Tundra, because it is very cold and dry (1); the vegetation is low, with mosses, lichens, grasses and small shrubs and no trees (1). (b) Station R is a desert, because it is hot and very dry (1), and a plant there may have a thick waxy cuticle, spines, or deep or wide roots (1; any one adaptation).

Final answer

(a) Tundra; no trees, with mosses, lichens and low shrubs. (b) Desert; a plant feature such as a thick waxy cuticle, spines or deep roots.

Advanced practice

Question C1

[7 marks]

Two barnacle species, X and Y, were counted in 100 cm^2 quadrats at different heights on a rocky shore. The time each level is uncovered by the tide each day is also given. Height above chart datum / m | 0.4 | 1.0 | 1.6 | 2.2 | 2.8; Species X / per 100 cm^2, 2, 9, 14, 6, 1; Species Y / per 100 cm^2, 18, 11, 4, 0, 0; Time uncovered / h per day, 2, 5, 9, 14, 18 (a) Describe the distribution of species Y. [2] (b) Suggest two abiotic reasons why few barnacles of either species live at the highest levels. [2] (c) Outline an experiment to test whether time out of water limits the distribution of Y. [3]

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(a) Species Y decreases with height from 18 per quadrat at 0.4 m (1) to zero at 2.2 m and 2.8 m (1). (b) The higher levels are uncovered for 14 to 18 h per day (1), so barnacles dry out, cannot feed while the water is away and meet larger temperature changes (1). (c) Keep samples of Y on identical tiles in tanks, drained for different numbers of hours per day (1); the independent variable is time out of water and the dependent variable is the number surviving after a set time (1); keep the species, size, water temperature and tank volume the same and use several tiles per treatment (1).

Final answer

(a) Y falls from 18 at 0.4 m to 0 at 2.2 m and above. (b) Higher levels are uncovered for longer (14 to 18 h), so the barnacles dry out, cannot feed and meet greater temperature extremes. (c) Independent: time out of water; dependent: survival or number alive; controls and repeats.

Question C2

[7 marks]

Seeds of two plant species were incubated at six temperatures. The table gives the mean percentage that germinated. Temperature / ^\circC | 10 | 15 | 20 | 25 | 30 | 35; Species M / % germinated, 10, 38, 82, 90, 41, 0; Species N / % germinated, 0, 4, 21, 55, 88, 35 (a) Calculate the percentage change in germination of species N between 25 ^\circC and 30 ^\circC. [1] (b) Estimate the optimum temperature for each species from the data. [2] (c) State the range of temperatures over which species M germinated. [1] (d) Deduce which species would cope better with a warmer climate, and state a limit of the data. [3]

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(a) (88-55)/55\times100=60\,\% (1). (b) The highest germination for M is at 25 ^\circC and for N at 30 ^\circC (1 each, 2). (c) M germinated between 10 and 30 ^\circC and did not at 35 ^\circC (1). (d) Species N would cope better with warming (1), because its optimum is higher and it still germinates at 35 ^\circC (1); the six temperatures are 5 ^\circC apart, so the exact optimum and limits are uncertain (1).

Final answer

(a) 60 %. (b) M 25 ^\circC; N 30 ^\circC. (c) M germinated from 10 to 30 ^\circC and not at 35 ^\circC. (d) N, since its optimum is 30 ^\circC and it germinates at 35 ^\circC when M does not; the data cover only six temperatures, so the exact limits and optimum are not known.

Question C3

[6 marks]

Stenocara beetles live in the Namib Desert, where fog drifts in from the sea at dawn but rain is almost unknown. On a foggy morning the beetle climbs a dune crest, faces the wind and tips its body head-down. Droplets collect on small bumps on its back and run towards its mouth. (a) Classify the two adaptations as structural, behavioural or physiological. [2] (b) Explain how the bumps and the posture provide the beetle with water. [2] (c) Suggest one further feature of its body surface that would help it survive, and explain why. [2]

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(a) Climbing the dune and tilting the body are behavioural (1); the bumps are structural (1). (b) The bumps give surfaces on which fog droplets condense and merge (1), and the head-down posture lets gravity move the water to the mouth, so the beetle gets water without drinking (1). (c) A waxy cuticle on the body reduces evaporation (1), so less of the collected water is lost by evaporation in the dry desert air (1). Accept any structural feature that reduces water loss, with its effect.

Final answer

(a) Climbing and tilting: behavioural; the bumps: structural. (b) The bumps give a surface on which fog droplets collect and grow, and the head-down posture lets the water run to the mouth. (c) A waxy cuticle on the rest of the body to reduce evaporation.

Question C4

[6 marks]

A team measured five sites as possible places to establish coral. Reef formation needs temperatures of 23 to 34 ^\circC, shallow clear water and sea water of normal salinity, about 35 parts per thousand. Site | P | Q | R | S | T; Depth / m, 6, 6, 5, 45, 6; Temperature / ^\circC, 28, 32, 28, 27, 28; Visibility / m, 20, 18, 2, 25, 20; Salinity / parts per thousand, 35, 35, 35, 35, 12 (a) Identify the three sites that are unsuitable and the condition that each fails. [3] (b) Recommend P or Q, with a reason. [1] (c) Predict the effect of a 3 ^\circC rise in sea temperature at site Q, using the idea of bleaching. [2]

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(a) Site R fails on clarity, because visibility is only 2 m so little light reaches the algae (1); site S fails on depth, because 45 m is too deep for light (1); site T fails on salinity, because 12 is far below 35 (1). (b) Site P (1), because its temperature of 28 ^\circC is well inside the range, while Q at 32 ^\circC is near the upper limit. (c) At Q the temperature would reach 35 ^\circC, above 34 ^\circC (1), so the corals would be stressed and expel their zooxanthellae (bleaching), losing their food supply and risking death (1).

Final answer

(a) R: cloudy (visibility 2 m); S: too deep (45 m); T: salinity 12, too low. (b) Site P, with 28 ^\circC, in the middle of the range. (c) At Q 32 ^\circC rises to 35 ^\circC, above 34 ^\circC: stress, bleaching, expulsion of the algae and loss of food.

Connected questions

Question X1

Connected questions[6 marks]

B4.1 + D2.3 + B3.2. Salt-marsh plants grow in soil that is flooded with sea water twice a day. A halophyte is compared with a freshwater plant placed in the same soil. (a) Explain, using osmosis, why the freshwater plant wilts in the salty soil. [2] (b) Suggest how the halophyte can still take in water from the same soil. [2] (c) Explain how sunken stomata reduce water loss by transpiration. [2]

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(a) The soil solution is more concentrated than the sap of the freshwater plant, so it is hypertonic (1); water leaves the root cells by osmosis and the cells lose turgor, so the plant wilts (1). (b) The halophyte accumulates solutes so that its root cells have a higher solute concentration than the soil water (1), so water still enters its roots by osmosis (1). (c) A sunken stoma holds a pocket of humid still air, so the difference in water vapour concentration between the leaf and the air is smaller (1) and less water is lost by transpiration (1).

Final answer

(a) Salty soil water is more concentrated than the root sap, so water leaves the roots by osmosis and the plant wilts. (b) The halophyte makes its root sap more concentrated than the soil water, so water enters by osmosis. (c) Humid air trapped in the pit lowers the difference in vapour concentration, so less water diffuses out.

Question X2

Connected questions[6 marks]

B4.1 + C1.3 + C4.1. Reef-building corals contain symbiotic algae. A heat wave warms the water at a reef by 3 ^\circC for several weeks, and the corals turn white. (a) Explain why reef-building corals need shallow, clear water. [2] (b) Explain why the corals turn white and may die. [2] (c) Predict the effect on the diversity of fish on the reef, with a reason. [2]

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(a) The algae need light for photosynthesis (1), and the sugars they make feed the coral polyp, so reefs need shallow, clear water (1). (b) The stressed coral expels its algae, which is bleaching (1), so the coral loses its main food source and may die (1). (c) Dead coral gives fewer shelters and less food (1), so fewer fish species can live there and species richness falls (1).

Final answer

(a) The algae photosynthesise and pass sugars to the polyp; they need light, so shallow clear water. (b) The corals lose their algae (bleaching), so lose their food and may die. (c) Fewer corals mean fewer shelters and food sources, so fish diversity falls.

Question X3

Connected questions[6 marks]

B4.1 + D4.3 + B4.2. In the Arctic the mean temperature has risen and the growing season is longer. Shrubs are spreading north into the tundra. (a) State one abiotic change that allows the shrubs to spread. [1] (b) Explain why the spread of shrubs could reduce the population of tundra mosses. [2] (c) Thawing permafrost releases methane and carbon dioxide. Explain how this could increase warming. [3]

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(a) A rising temperature, or a longer growing season (1). (b) Shrubs overshadow the low moss (1), so the moss loses light, which it needs for photosynthesis, and the two compete for the same resources, so the moss population falls (1). (c) Thawing permafrost releases methane and carbon dioxide (1), which are greenhouse gases that trap heat (1), causing further warming and more thawing, so the effect is amplified: a positive feedback (1).

Final answer

(a) Temperature (rising) or length of the growing season. (b) Taller shrubs shade the moss and use the same resources, so the moss loses light and its niche is reduced. (c) Thawing permafrost releases methane and carbon dioxide, which are greenhouse gases that trap more heat, so more warming and more thawing: a positive feedback.

Question X4

Connected questions[6 marks]

B4.1 + A4.1 + A3.2. Cacti in American deserts and euphorbias in African deserts both have thick, spiny, water-storing stems, yet they belong to different plant families. (a) State what is meant by convergent evolution. [1] (b) Explain how natural selection could produce the same stem features in both groups. [3] (c) Outline evidence that cacti and euphorbias are not closely related. [2]

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(a) Convergent evolution is the evolution of similar features in unrelated groups, because they face similar conditions (1). (b) In both deserts, water is scarce, so plants with water-storing stems and spines survived better (1) and left more offspring (1); the alleles for these features became more common over generations (1). (c) Differences in base sequences of DNA or proteins (1) and differences in flower structure (1) show that they are not closely related.

Final answer

(a) Similar features evolve in unrelated groups in similar conditions. (b) In both deserts plants with stem water stores and spines survived and left more offspring, so those alleles spread. (c) Differences in DNA or protein sequences or in flower structure show they are not closely related.

From the book

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

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

About the book