IGCSE Biology Paper 4 (Theory / Extended)

Topic 20: Human Influences on Ecosystems -- Challenge Exam 2

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0610

Instructions

Answer all questions in the spaces provided.
Show all working for calculations.
Use appropriate scientific terminology.
Your answers will be automatically graded when you submit.

Question Navigation

This paper covers the whole of Topic 20. Like a real Cambridge paper it ranges across every sub-topic — 20.1 food supply, 20.2 habitat destruction, 20.3 pollution and 20.4 conservation — and it mixes them inside single questions, so one question may begin with a farm and end with a fishery. All three Topic 20 papers do this; they differ in the angle they come at the material from, not in what they cover.

Question 1 — One Lake, Three Curves

Total: 12 marks

Fig. 1.1 shows three measurements made in a lake every month for one year. Curve X is the nitrate concentration, curve Y is the mass of algae and curve Z is the dissolved oxygen concentration. Each is plotted as a percentage of the highest value recorded for it during the year.

(a) [3]

State the month in which X reaches its maximum, the month in which Y reaches its maximum, and the month in which Z reaches its minimum.

Model Answer — 1(a)

X reaches its maximum in month 3 [1]

Y reaches its maximum in month 5 [1]

Z reaches its minimum in month 6 [1]

⚠ If you missed marks here: Read from the peak of the curve down to the axis with a ruler, and give the month, not the value. Marks are lost here by giving the height of the peak instead of its position, and by reading Z as month 5 because the eye is still on curve Y.

1 mark — X maximum in month 3 1 mark — Y maximum in month 5 1 mark — Z minimum in month 6

(b) [3]

Calculate the percentage decrease in Z between month 1 and month 6. Show your working. Then calculate the mean rate of increase of Y between month 3 and month 5.

Model Answer — 1(b)

Z falls from about 88 % to about 18 %, a fall of about 70 [1]

70 ÷ 88 × 100 = about 80 % decrease (accept 76–83 %) [1]

Y rises from about 58 % to about 98 % in 2 months, so the mean rate is (98 − 58) ÷ 2 = about 20 % of the maximum per month [1]

⚠ If you missed marks here: A percentage decrease is always divided by the starting value, not by the final one and not by 100. For the rate, the unit is part of the answer: a number with no “per month” after it is not a rate. Reading tolerances are generous, but the working must be visible.

1 mark — reads Z as about 88 at month 1 and about 18 at month 6 1 mark — percentage decrease of about 80 % with working shown 1 mark — rate of about 20 units per month for Y, with working

Explain why the three curves change in the order shown in Fig. 1.1.

Model Answer — 1(c)

X rises first because nitrate ions are washed into the lake from the land; the algae then use those ions, so Y rises after X and X falls as it is taken up [1]

Y falls after month 5 as the algae die; the decomposers that break down the dead algae increase in number [1]

Z falls last and reaches its minimum after Y has peaked, because the decomposers use the dissolved oxygen in aerobic respiration [1]

⚠ If you missed marks here: The order is the answer: nitrate, then algae, then oxygen, each one causing the next. A common error is to say the oxygen falls because the algae block out the light — the fall is caused by decomposers respiring, and it comes after the algae have died, which is why the minimum in Z is later than the peak in Y.

1 mark — nitrate arrives first and is then used by the algae, so Y follows X 1 mark — the algae die and decomposer numbers increase 1 mark — decomposers use dissolved oxygen in aerobic respiration, so Z falls last

(d) [3]

A student concludes from Fig. 1.1 that “the algae poisoned the fish”. Give two reasons why the data do not support this conclusion, and state what the data do show.

Model Answer — 1(d)

no measurement of fish was made at all, so the graph contains no evidence about fish [1]

the graph shows only that the three measurements change one after another; changing together in time does not by itself prove that one causes another [1]

what the data do show is that the dissolved oxygen falls after the mass of algae peaks, which is consistent with organisms dying from a lack of oxygen rather than from a poison [1]

⚠ If you missed marks here: Two separate criticisms are wanted, and “the graph does not show fish” twice in different words counts once. The third mark rewards saying what the data can support — examiners like a candidate who corrects a conclusion instead of just rejecting it.

1 mark — no data on fish were collected 1 mark — the data show a sequence in time, which is not proof of cause 1 mark — the data show oxygen falling after the algae peak, consistent with a lack of oxygen

Question 2 — Thirty Kilometres Downstream

Total: 12 marks

Fig. 2.1 shows the dissolved oxygen concentration measured along a river. A pipe carrying water that has drained from farmland treated with fertiliser discharges into the river at 4 km. The river flows from left to right, and its width and depth do not change.

(a) [3]

State the dissolved oxygen concentration at 0 km, the minimum concentration reached, and the distance downstream at which that minimum occurs.

Model Answer — 2(a)

at 0 km the concentration is about 10 mg per dm³ (accept 10.0–10.4) [1]

the minimum is about 1 mg per dm³ [1]

and it occurs at about 12 km [1]

⚠ If you missed marks here: Always give the unit when a graph provides one; a bare number is not a concentration. Take the minimum from the lowest point of the curve, not from the point where it crosses the dashed line.

1 mark — about 10 mg per dm³ at 0 km 1 mark — minimum of about 1 mg per dm³ 1 mark — minimum occurs at about 12 km

(b) [3]

Calculate the percentage decrease in dissolved oxygen between the pipe at 4 km and the minimum. Show your working. Then use the graph to estimate the length of river along which most fish cannot survive.

Model Answer — 2(b)

at 4 km the concentration is about 9.2 mg per dm³ and the minimum is about 1.0, a fall of about 8.2 [1]

8.2 ÷ 9.2 × 100 = about 89 % (accept 85–92 %) [1]

the curve is below 4 mg per dm³ from about 7.5 km to about 18 km, so about 10 km of river (accept 9–12 km) [1]

⚠ If you missed marks here: The second calculation is done by reading across from the dashed line at 4 mg per dm³ to both places where the curve crosses it, then subtracting the two distances. Students often give only the point where the curve first drops below the line, which answers a different question.

1 mark — reads about 9.2 at 4 km and about 1.0 at the minimum 1 mark — percentage decrease of about 89 % with working 1 mark — about 10 km of river below 4 mg per dm³

Explain why the dissolved oxygen concentration rises again after 12 km.

Model Answer — 2(c)

no more ions are added downstream, and those present have been used up by the producers or diluted, so no further growth is stimulated [1]

most of the dead material has already been broken down, so the number of decomposers and the rate at which they use oxygen both fall [1]

oxygen re-enters the water from the air, and from photosynthesis by the plants living in the river, so the concentration returns towards its original value [1]

⚠ If you missed marks here: The recovery has two halves: less oxygen being used, and oxygen coming back in. Most answers give only the first. The river also carries the pollution away from any one point, which is why the same spot improves over distance — not because the pollution has disappeared.

1 mark — no further ions added / ions used up or diluted 1 mark — fewer decomposers, so less oxygen is used 1 mark — oxygen re-enters from the air and from photosynthesis

(d) [3]

The readings were all taken at 10 am on one day. Describe two improvements to the method that would make the conclusion more reliable, giving a reason for each, and state one variable that had to be kept the same along the river.

Model Answer — 2(d)

repeat the readings on several days and at several times of day, because the dissolved oxygen changes as the rate of photosynthesis changes through the day [1]

take readings at the same points in a similar river with no pipe, as a control, so that the effect of the pipe can be separated from the natural pattern of the river [1]

a variable kept the same: the width and depth of the river, or the temperature of the water, which affects how much oxygen dissolves [1]

⚠ If you missed marks here: “Repeat the experiment” without a reason earns nothing at this level. Say what changes between repeats and why that matters. The control here is a second river or the water upstream of the pipe; without one, any pattern could simply be how the river always behaves.

1 mark — repeat on several days or times, because oxygen varies through the day 1 mark — use a control river or readings upstream of the pipe 1 mark — names a controlled variable such as depth, flow or temperature

Question 3 — Two Areas of Sea, Twenty Years

Total: 12 marks

Fig. 3.1 shows the mass of fish in two areas of sea, P and Q, each fished for 20 years. Both areas started with the same mass of fish and were fished by the same method. The only difference between them was the mass of fish taken each year.

(a) [2]

Describe the difference between the change in the stock of area P and the change in the stock of area Q over the 20 years.

Model Answer — 3(a)

in P the stock varies from year to year between about 73 and 81 thousand tonnes but shows no overall fall, ending at about the same value as it started [1]

in Q the stock falls continuously from 80 to about 2 thousand tonnes [1]

⚠ If you missed marks here: Quote figures from the axes; “P stays the same and Q goes down” is a description of the shape, not of the data, and is worth one mark at most. Note also that P does not stay exactly constant — saying that it varies but does not fall overall is the more accurate answer.

1 mark — P fluctuates between about 73 and 81 thousand tonnes with no overall fall 1 mark — Q falls from 80 to about 2 thousand tonnes

(b) [3]

Calculate the percentage decrease in the stock of area Q over the 20 years, and its mean rate of decrease. Show your working.

Model Answer — 3(b)

(80 − 2) = 78 thousand tonnes lost [1]

78 ÷ 80 × 100 = 97.5 % (accept 96–98 %) [1]

78 ÷ 20 = 3.9 thousand tonnes per year [1]

⚠ If you missed marks here: Divide by the starting mass of 80, not by the final mass of 2, which would give a nonsensical figure of nearly 4000 %. The mean rate is the total change divided by the total time, and it needs the unit “per year” to be a rate at all.

1 mark — fall of 78 thousand tonnes identified 1 mark — 97.5 % decrease with working 1 mark — mean rate of 3.9 thousand tonnes per year

Between years 0 and 8 the stock of Q fell by 48 thousand tonnes. Between years 12 and 20 it fell by 13 thousand tonnes. A fisherman says that the smaller fall shows that the stock is recovering. Explain why the data do not support this.

Model Answer — 3(c)

the two rates are 48 ÷ 8 = 6 and 13 ÷ 8 = about 1.6 thousand tonnes per year, so the fall is slower [1]

but the stock is still falling, not rising, so it is not recovering [1]

the fall is slower only because there are far fewer fish left to catch, and with fewer adults fewer eggs are produced, so replacement is slower still [1]

⚠ If you missed marks here: A slowing decline is still a decline. Work out both rates before commenting — the mark for the comparison is separate from the mark for the judgement. The last mark is the one that shows understanding: a stock that has crashed replaces itself slowly, which is why recovery is not automatic once fishing eases.

1 mark — calculates the two rates, about 6 and about 1.6 thousand tonnes per year 1 mark — the stock is still falling, so it is not recovering 1 mark — fewer fish remain to be caught, and fewer adults produce fewer offspring

(d) [4]

Using the definition of a sustainable resource, explain which of the two areas is being fished sustainably, and state what would have to change in the other area.

Model Answer — 3(d)

a sustainable resource is one produced as rapidly as it is removed, so that it does not run out [1]

area P is sustainable: the mass taken each year is matched by the mass replaced by reproduction, so the stock does not fall [1]

area Q is not: more is taken each year than is replaced, which is overharvesting [1]

in Q the mass taken each year would have to be reduced below the rate of replacement — by a quota, a closed season or a protected area — and the stock monitored to check that it rises [1]

⚠ If you missed marks here: Notice that the graph never tells you how many tonnes were caught in either area; it tells you what happened to the stock, and you infer the catch from that. Marks are lost by writing that P is sustainable “because less was caught” — the amount alone decides nothing, only the amount compared with the rate of replacement.

1 mark — definition given as a comparison of two rates 1 mark — P is sustainable because catch matches replacement 1 mark — Q is overharvested because catch exceeds replacement 1 mark — a named change: reduce the catch below replacement, with a quota, closed season or protected area

Question 4 — Two Nets, One Week

Total: 12 marks

Fig. 4.1 shows two fishing nets, R and S, used in the same area of sea by boats of the same size. The shaded fish are adults and the pale fish are young fish that have not yet bred. Table 4.1 shows what each net caught in one week.

net	mass of adult fish caught / kg	mass of young fish caught / kg
R	480	320
S	460	40

(a) [2]

State two differences between the catches of net R and net S that can be seen in Fig. 4.1.

Model Answer — 4(a)

net R has the smaller mesh, of 40 mm, and net S the larger, of 95 mm [1]

net R holds both the adults and the young fish, whereas in net S the young fish pass through the mesh and escape [1]

⚠ If you missed marks here: Read the labels on the diagram: the mesh sizes are printed, so quote them. A difference must mention both nets — “net S lets small fish through” is only half a comparison until you say what net R does with them.

1 mark — net R mesh 40 mm, net S mesh 95 mm 1 mark — R retains the young fish, S allows them to escape

(b) [3]

Using Table 4.1, calculate the percentage of the total catch that was young fish for each net. Show your working. Then state the difference between the two nets in the mass of adult fish caught.

Model Answer — 4(b)

net R: 320 ÷ 800 × 100 = 40 % [1]

net S: 40 ÷ 500 × 100 = 8 % [1]

net S caught 480 − 460 = 20 kg fewer adults, which is only about 4 % less [1]

⚠ If you missed marks here: The denominator is the total catch of that net, which you have to work out first by adding the two columns — 800 kg and 500 kg. Dividing by 480 or 460 instead is the usual slip. The third mark matters for the argument that follows: the larger mesh costs the boat very little in adult fish.

1 mark — 40 % for net R with working 1 mark — 8 % for net S with working 1 mark — net S caught 20 kg fewer adults / about 4 % less

Explain how the use of net S rather than net R conserves the fish stock.

Model Answer — 4(c)

the young fish have not yet bred, and they pass through the larger mesh instead of being caught [1]

so they survive to reach breeding age and reproduce, replacing the fish that were taken [1]

the rate of replacement then matches the rate of removal, so the stock does not fall and can be fished year after year [1]

⚠ If you missed marks here: The chain is: escape, breed, replace. Missing out the middle step leaves “small fish escape so there are more fish”, which does not explain anything. Controlled net type and mesh size is one of the six named methods of conserving fish stocks; the others are education, closed seasons, protected areas, quotas and monitoring.

1 mark — young fish that have not bred escape through the larger mesh 1 mark — they survive to breed and replace the fish taken 1 mark — removal then matches replacement, so the stock does not fall

(d) [4]

A government makes net S compulsory and wants to find out whether the rule is working. Describe how the fish stock could be monitored, and state two variables that must be kept the same if the catches of different years are to be compared.

Model Answer — 4(d)

take sample catches at regular intervals over several years, recording the mass caught and the number of fish in each size or age group [1]

a rising proportion of young fish surviving to become adults, or a rising mass caught for the same effort, shows that the rule is working [1]

keep the same fishing effort: the same number of boats fishing for the same number of hours [1]

and sample in the same area at the same time of year, using the same net and mesh size [1]

⚠ If you missed marks here: Monitoring means measuring the same thing repeatedly and comparing, so a single survey after the rule is introduced proves nothing. The variable most often forgotten is fishing effort: if twice as many boats fish, a larger catch says nothing about the size of the stock.

1 mark — regular sampling over several years, recording mass and size or age 1 mark — states what result would show the rule is working 1 mark — same fishing effort: boats and hours 1 mark — same area, season and equipment

Question 5 — Four Plots of Wheat

Total: 10 marks

A farmer treated four plots of wheat in different ways and measured the yield of each. The results are shown in Table 5.1.

treatment of the plot	yield of wheat / tonnes per hectare
no treatment	3.2
fertiliser only	5.6
insecticide only	4.0
fertiliser and insecticide	6.8

(a) [3]

Calculate the percentage increase in yield produced by fertiliser alone, and the percentage increase produced by fertiliser and insecticide together. Show your working.

Model Answer — 5(a)

(5.6 − 3.2) ÷ 3.2 × 100 [1]

= 75 % [1]

(6.8 − 3.2) ÷ 3.2 × 100 = 112.5 % [1]

⚠ If you missed marks here: A percentage increase is the change divided by the original, so the untreated yield of 3.2 is the denominator both times. Dividing by the new yield gives 43 % and 53 %, which are the commonest wrong answers to this style of question.

1 mark — working shown for the first calculation 1 mark — 75 % 1 mark — 112.5 %

(b) [3]

Explain why the two treatments together give a higher yield than either of them alone.

Model Answer — 5(b)

the fertiliser supplies mineral ions such as nitrate, used to make amino acids and proteins, so the plants grow larger and produce more grain [1]

the insecticide kills insect pests, so less of the crop is eaten or damaged and more of what grows is harvested [1]

the two act on different factors that were limiting the yield, so their effects add together [1]

⚠ If you missed marks here: Say what each chemical does, then why the two are not doing the same job. The last mark is the one that answers the question actually asked — well-fed plants are still eaten by insects, and protected plants still cannot grow without mineral ions.

1 mark — fertiliser supplies mineral ions for growth 1 mark — insecticide kills pests, so less crop is lost 1 mark — the two remove different limiting factors, so the effects add

State two variables that must be kept the same for this comparison to be valid.

Model Answer — 5(c)

the same variety of wheat, and the same area of plot [1]

the same soil type and the same weather, water supply and date of harvesting [1]

⚠ If you missed marks here: Give variables that could realistically differ between four plots on one farm, and be specific: “the same conditions” is too vague to score. Note that the treatment itself is the independent variable, so it is not one of the things kept the same.

1 mark — same variety of wheat / same plot area 1 mark — same soil, water, weather or harvest date

(d) [2]

Only one plot was used for each treatment, in one year. Explain why the farmer should not yet conclude that fertiliser and insecticide together are the best treatment for his farm.

Model Answer — 5(d)

there was only one plot per treatment and no repeats, so the differences could be caused by variation between the plots, such as differences in the soil, rather than by the treatment [1]

the trial ran for only one year, and the weather and the number of pests differ between years, so it should be repeated before the result is trusted [1]

⚠ If you missed marks here: Two distinct criticisms are wanted: no repeats within the year, and only one year. “It is not accurate” scores nothing — name what could have caused the difference instead of the treatment.

1 mark — no repeats, so plot-to-plot variation could explain the result 1 mark — only one year, so the trial should be repeated in other conditions

Question 6 — Chicken, Two Ways

Total: 12 marks

Table 6.1 compares two systems used to produce 1000 kg of chicken meat. In the indoor system the birds are kept at high density in a heated building; in the outdoor system they range freely over grassland.

	indoor system	outdoor system
mass of food eaten / kg	1800	2600
area of land used / hectares	0.4	5.0
time taken to reach mass for sale / days	42	81

(a) [3]

Calculate the mass of food eaten per kilogram of meat produced in each system, and calculate how many times more land the outdoor system uses. Show your working.

Model Answer — 6(a)

indoor: 1800 ÷ 1000 = 1.8 kg of food per kg of meat [1]

outdoor: 2600 ÷ 1000 = 2.6 kg of food per kg of meat [1]

5.0 ÷ 0.4 = 12.5 times more land [1]

⚠ If you missed marks here: Keep the units attached: the first two answers are masses of food per unit mass of meat, and the third is a ratio with no unit at all. Dividing the wrong way round gives 0.08 times, which should look obviously wrong against a figure you can see in the table.

1 mark — 1.8 kg of food per kg of meat, indoor 1 mark — 2.6 kg of food per kg of meat, outdoor 1 mark — 12.5 times more land, with working

(b) [4]

Using Table 6.1 and your own knowledge, discuss the advantages and disadvantages of intensive livestock production. At least one of each is required.

Model Answer — 6(b)

advantage: less food is needed per kilogram of meat, because the birds move less and are kept warm, so less energy is used in movement and in maintaining body temperature [1]

advantage: far less land is needed, and the birds reach the mass for sale in about half the time, so more meat is produced more cheaply and more people can be fed [1]

disadvantage: the animals are crowded, so disease spreads quickly between them and medicines may have to be used routinely [1]

disadvantage: the animals have restricted movement, which raises questions about their welfare, and a large amount of waste is produced in one place, which may pollute nearby water [1]

⚠ If you missed marks here: This objective is symmetric and is marked symmetrically: an answer that only attacks intensive production, however well argued, is capped at half. The advantages here are real and are visible in the table — less food, less land, less time. Use the figures rather than asserting the advantages in general terms.

1 mark — advantage: less food needed per kg of meat, with an energy reason 1 mark — advantage: less land or faster growth, so cheaper meat / more people fed 1 mark — disadvantage: crowding, so disease spreads quickly 1 mark — disadvantage: welfare of the animals or waste polluting nearby water

A country with a large number of cattle finds that the methane released is a concern. Explain how this affects the atmosphere, and state one other source of methane.

Model Answer — 6(c)

methane is one of the two gases named as adding to the enhanced greenhouse effect [1]

a higher concentration returns more long-wave radiation to the Earth’s surface, so the surface is warmer and the climate changes [1]

another source of methane is flooded rice fields, or the decomposition of organic waste [1]

⚠ If you missed marks here: Do not describe methane as a poison or as damaging the air chemically. Its effect is entirely about radiation returned to the surface. And the effect is an enhancement of something natural, not the creation of something new.

1 mark — methane contributes to the enhanced greenhouse effect 1 mark — more long-wave radiation returned, so the surface warms / climate changes 1 mark — another named source, for example flooded rice fields or decomposing waste

(d) [2]

Explain why growing wheat for people to eat feeds more people per hectare than growing the same wheat to feed cattle.

Model Answer — 6(d)

eating the wheat directly involves one energy transfer fewer [1]

at each transfer energy is lost in respiration as heat, in faeces, in urine and in the parts of the animal that are not eaten, so only a small fraction of the energy in the wheat reaches a person through the cattle [1]

⚠ If you missed marks here: Count the transfers first, then name at least one specific loss. “Energy is lost” on its own is half an answer. It is also worth knowing the fair reply to this argument: some land is too steep, poor or dry to grow crops on but will still support grazing animals.

1 mark — one fewer energy transfer when the crop is eaten directly 1 mark — names a specific loss at the extra transfer, such as respiration, faeces or parts not eaten

Question 7 — Beetles in Three Parts of a Valley

Total: 10 marks

A student set ten traps in each of three areas of a valley and identified every beetle caught. Table 7.1 shows the results.

area	description	number of different species of beetle	total number of beetles caught
A	forest never cleared	34	410
B	forest cleared 2 years ago	11	380
C	forest cleared 20 years ago	19	395

(a) [2]

State what is meant by biodiversity, and identify which column of Table 7.1 measures it.

Model Answer — 7(a)

biodiversity is the number of different species that live in an area [1]

it is measured by the column headed “number of different species of beetle”, not by the total number of beetles caught [1]

⚠ If you missed marks here: The table is built to catch the usual mistake. The total number of beetles is a measure of how many individuals there are, which is population size; biodiversity counts how many different kinds there are. The two columns move independently, which is exactly the point.

1 mark — biodiversity is the number of different species in an area 1 mark — identifies the number of different species column, not the total

(b) [3]

Calculate the percentage decrease in biodiversity between area A and area B. Show your working. Then describe what the comparison of areas B and C shows.

Model Answer — 7(b)

(34 − 11) ÷ 34 × 100 [1]

= about 68 % decrease [1]

area C has more species than B but still fewer than A, so some species have returned during the 20 years since clearance but the area has not recovered fully [1]

⚠ If you missed marks here: Use the species column for both parts — a percentage worked out from the totals answers a different question. For the comparison, say both things: recovery has happened, and it is incomplete. Only one of the two is usually written down.

1 mark — working shown: (34 − 11) ÷ 34 × 100 1 mark — about 68 % 1 mark — C is higher than B but lower than A, so recovery is partial

Describe how the traps should be used so that the three areas can fairly be compared.

Model Answer — 7(c)

the same number of traps in each area, of the same type and with the same bait [1]

left for the same length of time, and at the same time of year and in similar weather [1]

placed at randomly chosen positions within each area, and enough of them used for the sample to represent the whole area [1]

⚠ If you missed marks here: Fair comparison means everything except the area itself is the same. The mark that is usually missed is the one for random positioning: traps put where the student expects to find beetles will always find them, and the result then describes the choice of position rather than the area.

1 mark — same number, type and bait of trap 1 mark — same length of time and same time of year 1 mark — positions chosen at random / enough traps to be representative

(d) [2]

The total number of beetles caught is similar in all three areas. Explain why this does not mean that the three areas are equally rich in wildlife.

Model Answer — 7(d)

the total counts individuals, not species, so it says nothing about how many different kinds of beetle are present [1]

a few species can become very numerous while many others are lost, keeping the total high while the biodiversity falls [1]

⚠ If you missed marks here: This is the same idea as part (a), tested the other way round, and it is worth being able to state in one sentence: the same number of individuals can be shared among very different numbers of species.

1 mark — the total counts individuals, not different species 1 mark — a few species may become numerous while many are lost

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