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1. Moose are large herbivorous animals.

(a) In a study of one population of moose, 72 animals were trapped and marked with ear tags. They were then released. One month later, fieldworkers examined 120 moose and found that 14 of these had ear tags.

Use these figures to calculate the size of the moose population. Show your working.

Answer:............................(2)

(b) Isle Royale is a large island in Lake Superior. Moose first colonised this island in 1900. At the time they had no predators on the island. Wolves, which are predators of moose, were introduced to the island in 1950. The graph shows the moose population from 1900.

Isla nd co lo n ise db y m o o se

W o lv e sin tro d u ced

3 00 0

2 00 0

1 00 0

019 0 0 1 9 1 0 1 9 2 0 1 9 3 0 1 9 4 0 1 9 50 1 9 6 0 1 9 7 0

D ate

M o o sep o p u la tio n

Woodford High School 1

(i) Suggest an explanation for the changes in the moose population before the introduction of the wolves.

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(ii) Describe the effect of introducing wolves on the moose population.

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(Total 5 marks)

2. The diagram represents annual energy flow, in kJ m–2 year–1, through an aquatic ecosystem.

D ec o m p o sers

H erb iv o re s

P ro d u cers

S u nlig h t7 1 4 0 0 0 0

1 4 1 5 0

8 7 4 0 0

7 2 5 0

1 7 80 0

lo s t9 5 00

lo s t7 0 5 2 6 0 0


(a) Give two ways in which energy is lost between producers and herbivores.

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(2)

(b) Calculate the percentage of available sunlight trapped by the producers.Show your working.

Answer..............................(1)

(c) What is the importance of decomposers to the producers?

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(Total 4 marks)


3. Heather is a woody plant found in many moorland areas on higher ground in Britain. Grouse are chicken-sized birds which feed on the young shoots of heather. On many moorlands, the heather is managed to produce better conditions for grouse. This involves burning large areas. After burning, heather goes through the cycle of growth described below.

Pioneer phase0-7 years

New plants grow either fromseed or from stem basesundamaged by fire.

Building phase7-15 years

Phase of rapid growthproducing dense cover ofheather plants.

Mature phase15 - 25 years

Plants still growing but gapsstart to appear in the canopy.

Degenerate phase25+ years

Branches break. Central partof plant dies although outerbranches may remain alive.

Table 1 summarises some data collected from a study of heather plants of different ages.

Age/years Totalbiomass/g m2

Biomass of greenshoots/g m–2

Percentage cover Light at ground levelas percentage of light

in open

5 287 78 10 84

10 1508 203 85 5

20 1924 208 76 22

30 1043 n/a 34 56

Table 1

Key:n/a = figures not available


(a) To determine the total biomass, quadrats were used to collect the heather samples.Describe the procedure you would have used to place these quadrats.

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(b) Describe how and explain why the diversity of plant species would be likely to change during the cycle of growth of the heather.

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(c) (i) Calculate the mean rate of increase in the biomass of green shoots between 10 and 20 years. Show your working.

Answer: ............................................(2)


(ii) Explain why the mean rate of increase in the biomass of green shoots decreases with age.

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(iii) Management of heather by burning allows an area of land to support a larger population of grouse than if the heather is unmanaged. Use the information in Table 1 to suggest why.

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Table 2 shows the nitrogen content of heather plants, leaf litter and the soil in plots of land at various stages after burning.

Time after Nitrogen content/kg ha–1

burning/years Heather plants Leaf litter Soil

3 59 28 5794

8 92 61 5952

15 192 58 5394

Table 2

(d) (i) Give one organic compound in which the nitrogen will be found in the heather plants.

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(ii) Give one ion in which the nitrogen will be found in the soil.

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(e) Explain how bacteria make nitrogen present in the leaf litter available to the heather plants.

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(f) It has been suggested that if heather moorland is to be managed by burning, this should take place before the heather is 15 years old. Use the information in Table 2 to suggest why.

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(Total 20 marks)


4. The diagram shows the arrangement of some of the proteins in a myofibril from a skeletal muscle. The myofibril is shown in the relaxed state.

Z lin e

A b a n d I b a n dH zo n e

S a rco m ere

(a) Name the protein found in the H zone.

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(b) When the muscle contracts, what happens to the width of

(i) the A band; .......................................................................................................

(ii) the I band? ........................................................................................................(2)


S (c) The distance between two Z lines in a myofibril is 1.6 m. Calculate the magnification of the diagram. Show your working.

Answer ........................................................................................................................(2)

(Total 5 marks)

5. Many athletes practise glycogen loading. They eat food high in carbohydrates for 3 – 4 days before an event. This produces an increase in the amount of glycogen stored in the liver and muscles.

In an investigation into the effects of glycogen loading on performance, two groups of athletes followed different diets before exercising. All the athletes exercised at the same rate until they were exhausted. This time was recorded. The concentration of muscle glycogen was measured before and after exercise. Some results are shown in the graph.

M ix edd ie t

H ig hca rb o h y d ra te

d ie t

1 00 0

80 0

60 0

40 0

20 0

0

C o n cen tra tio no f m u sc leg ly c o ge n /m m o l k gd ry m a ss

– 1

K e yB efo re ex erc iseA fte r ex erc ise

F ig u res in b rack e tsin d ica te m ean tim eto e x h au s tio n inm in ute s

(1 2 6 )

(1 8 9 )


(a) (i) Calculate the percentage increase in mean time to exhaustion of the athletes following a high carbohydrate diet compared with those following a mixed diet.Show your working.

Answer ..............................................(2)

(ii) Explain this increase in mean time to exhaustion.

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S (b) What do the bars (I) on the graph indicate about the values?

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(Total 5 marks)


6. Like many insects, honeybees cannot survive in very cold conditions. Before an individual bee can fly, the temperature of its flight muscles must be above 15 °C. It achieves this with small, rapid contractions of the flight muscles, rather like shivering in a mammal.

Honeybees live as a colony in a hive. By clustering together in the centre of the hive they can keep the temperature of this area at a constant 34 to 35 °C. In this way the bees can survive in cold conditions.

In an investigation, the diameter of the cluster formed by one colony of bees was measured at various environmental temperatures. None of the bees died during the investigation. The results are shown in Figure 1.

3 6

3 3

3 0

2 7

2 4– 6 – 5 0 5 1 0 15

D iam eter o f th ec lu s te r / cm

E n v iro n m e nta l tem p era tu re / °C

Figure 1

(a) Describe and explain the relationship between the environmental temperature and the diameter of the cluster.

(2)


The table shows the results of a further investigation. It shows the relationship between the temperature in the centre of the cluster of bees and the sugar consumption of the bees.

Temperaturein the centre of cluster/°C

Mean sugar consumptionper bee/mg day–1

19 49

23 32

27 24

30 12

33 15

36 18

(b) Plot the data in the table as a suitable graph on the graph paper provided.(4)

(c) Explain the relationship between cluster temperature and sugar consumption:

(i) between 30 °C and 36 °C;

(ii) between 19 °C and 30 °C.(4)

(d) Use the data in the table to predict the effect of cluster temperature on carbon dioxide production by the bees. Explain your answer.

(2)


The graphs in Figures 2 and 3 show some aspects of the behaviour of bees.

Figure 2 shows the effect of an increase in carbon dioxide concentration in the hive on the number of bees which can be seen facing into the hive and fanning their wings vigorously.Figure 3 shows how the number of bees collecting water varies over the course of a day. These bees return to the hive and spread the water in tiny droplets over the internal surface of the hive.

1 5 0

1 2 0

9 0

6 0

3 0

00 1 2 3 4 5 6 7 8 9 1 0

C a rb o n d io x id e co n c en tra tio n / %

N u m b er o fbe es f an n in g

w in g s

Figure 2

4 0

3 6

3 2

2 8

2 4

2 0

1 6

1 2

8

4

0

4 0

3 5

3 0

2 5

2 0

Tem p era tu re / °C

N um b er o f b ees co llec tin g w a te r

N u m b er o fb ee s co llec tin g

w ate r

Te m p e ra tu re

Tim e of d ay9 .0 0 1 0 .0 0 11 .0 0 1 2 .0 0 1 3 .0 0 1 4 .0 0

Figure 3


(e) Use information from these graphs to explain how the internal temperature of the hive is prevented from rising on a hot day.

(3)

(f) Although all honeybees belong to the same species, there are many different varieties. One variety called the mountain bee is found at high altitudes in East Africa. It has longer ‘hair’ on its body than other African varieties of bee.

(i) Suggest why this longer ‘hair’ may be an advantage to bees living at high altitude.(2)

(ii) Explain how natural selection could have resulted in bees with longer ‘hair’.(3)

(Total 20 marks)

7. The graph shows changes in the concentration of some hormones during the oestrous cycle of a cow.

1 2

1 0

8

6

4

2

00 2 4 6 8 1 0 1 2 1 4 1 6 1 8 2 0 2 2

C o n ce n tra tio no f h o rm o n e/

a rb itra ry u n its

T im e /da y s

P ro g e ste ro n e

O es tro g en

L H

(a) Between which days is fertilisation most likely to be successful? Explain your answer.

Days ............................................................................................................................

Explanation .................................................................................................................

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(b) Did the cow become pregnant during this cycle? Give the evidence from the graph for your answer.

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(Total 3 marks)

8. The island of Surtsey was formed in 1963 by a series of volcanic eruptions. Biologists have studied the colonisation of this new island by plants and animals.

(a) The graph shows the number of species of plants found on the island each year for the first seven years of its existence.

1 6

1 4

1 2

1 0

8

6

4

2

0

N u m b ero f sp ec ie s

19 6 3 1 96 4 1 96 5 1 9 6 6 19 6 7 1 9 6 8 1 9 6 9 1 9 7 0Year


The development of the plant community on Surtsey illustrates the process of succession. Explain why:

(i) very few species could grow on the island in the first few years after it was formed;

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(ii) once some plants were growing, the rate at which new species could establish themselves increased rapidly.

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(b) Animals also colonised Surtsey. One conclusion drawn from studying them was that particular species of animals found here had broader ecological niches than they did on the nearby mainland. Suggest why.

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(Total 4 marks)


9. The graph shows the effect of applying different amounts of nitrogen fertiliser on the yield of wheat.

5 .0

4 .0

3 .0

2 .00 .0

0 2 0 40 6 0 8 0 1 0 0 12 0 1 4 0 1 6 0 1 8 0 2 0 0

Yie ld /to n n e s h a– 1

–1N itro g en fe r tilise r ap p lie d /k g h a

(i) Calculate the percentage increase in yield obtained when the amount of nitrogen fertiliser is increased from 50 to 100 kg ha–1. Show your working.

(2)


(ii) It would not be advisable for a farmer to apply more than 160 kg ha–1 of fertiliser to this wheat crop. Use the graph to explain why.

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(iii) Give two advantages of using an inorganic nitrogen fertiliser, rather than an organic fertiliser.

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(Total 6 marks)

10. Wheat is a cereal crop grown mainly for grain. Wheat straw consists of dead leaves and stems. An investigation was carried out into the effect of adding nitrogen-containing fertiliser on the yield of wheat.

M a ss o f fe rtilise ra d d ed /g m – 2

– 2Y ield / g m

G ra in4 7 85 4 96 2 2

0 .06 .3

5 0 .0

S traw2 4 52 8 43 2 4


(a) Does adding 50 g m–2 of nitrogen-containing fertiliser have a greater effect on the grain or the straw? Use suitable calculations based on the figures in the table to support your answer.

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(b) (i) When 6.3 g m–2 of nitrogen-containing fertiliser was added, the ratio of straw to grain was 0.517. Calculate the ratio of straw to grain when 50 gm–2 of nitrogen-containing fertiliser was added. Show your working.

(2)

(ii) Suggest why the grain to straw ratio may be important to a farmer.

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(c) Farmers remove weeds from land on which wheat is growing. Suggest why the yield of wheat grain is higher if weeds are removed at the beginning of the growing season than if they are removed later in the growing season.

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(Total 8 marks)

11. The diagram shows the annual transfer of energy in a forest ecosystem. The figures are inkJ m–2.

A bso rb e dby p ro d u cer s

4 .6 × 1 0

O rg an icm o lec u le s

4 4 0 9 0

H eat lo s s inresp ira tio n

2 0 0 0 0

P r im aryco n su m e rs

D e tr itu s1 5 3 3 0

S u n lig h t7 .1 × 1 0

M a te ria ls n o tc on su m ed

5 8 4 0

6

6

(a) Although the producers absorb 4.6 106 kJm–2, only about 1% of this is transferred to organic molecules. Suggest one reason for this.

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(b) Calculate the energy transferred to primary consumers in this ecosystem.

................................................... kJm–2

(1) (c) Nitrogen compounds in the detritus are broken down by bacteria to ammonium ions

)NH( 4

. Describe how ammonium ions are converted into a form that can be readilyabsorbed by the producers.

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(Total 5 marks)


12. The graph shows the blood glucose and glucagon concentrations in a healthy person during and after eating a meal.

1 2 0

1 0 0

8 0

1 2 0

11 0

1 0 0

9 0

M e al

G lu co se

G lu ca g o n

O neh ou r

b e fo rem eal

M e alb eg in s

Tw o h o u rsafte r m ea l

B lo o d g lu co sec o nc en tra tio n /

m g 1 0 0 cm

B lo o d g lu ca g o nco n ce ntra t io n /a rb itra ty u n its

– 3

(a) (i) From where is glucagon secreted?

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(ii) Explain the changes that occur in the glucagon concentration over the period on the graph.

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(b) Explain how the information shown in the graph provides an example of negative feedback.

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(Total 5 marks)

M/S

1. (a) Allow two marks for 617;allow one mark if arithmetical error in calculation; 2

(b) (i) Moose numbers rose in absence of predators / with plenty of food;depletion of food / spread of disease / climatic change /resulted in fall / death; 2

(ii) Caused the moose numbers to stabilise at lower level / equivalent; 1[5]

2.

(a) Respiratory loss;

Heat;

Undigested / gone to symbiont; max 2

(b) 87 400 ÷ 7 140 000 × 100 = 1.2% 1

(c) Supply of inorganic molecules / e.g. CO2 / nitrate / phosphate / minerals; 1[4]

3.

(a) Mark for principle of placing quadrats at random;marks for detail -(method of) marking area out with grid/using tapes;

acceptable method of generating random numbers; e.g.table/calculator; 3

(b) Any one correct change;reason for diversity level;explanation of how heather affects diversity; 3

(c) (i) Award two marks for 0.5 g m–2 yr–1;;award one mark for answer without unitsor 2.4 – 2.5 %; 2

(ii) Used for non photosynthesising/supporting organs;eaten by grouse; 2

(iii) Younger plants have more shoots/provide more food;Younger plants provide more cover;Getting rid of old (unproductive) plants; max. 2

(d) (i) proteins/nucleic acids/amino acids/ATP/nucleotides;

(ii) ammonium/nitrate/nitrite;


NH4 NO

–3 NO

–2 [Formula must correspond if given] 1

(e) Organic material/proteins to ammonia/ammonium compounds; 1by saprobiotic/putrefying bacteria;ammonia to nitrite;nitrite to nitrate;by nitrifying bacteria/named bacteria; max. 4

(f) In older stands, greater proportion of/more nitrogen in plants;more lost when burnt: 2

[20]

4. (a) Myosin; [Reject: any other protein] 1

(d) A band - stays the same width/ nothing;I band - becomes narrower/ shorter/ smaller; [Accept: Contracts] 2

(e) Correct working of 6.110048

/ or equivalent;= 30,000 ×; 2(Note: Allow 2 marks for correct answer even if no working shown but max1 mark if units of measurement given)

[5]

5.

(a) (i) 50% scores two marks;;If wrong answer, then 63/126 x 100 or 189/186 x 100 = 1 mark 2

(ii) More glycogen available at start of exercise / eq.;More energy released / ATP produced / respiration / glucoseproduced or for longer time; 2

(b) Bar indicates range of results / standard deviation / values aremeans (allow ‘averages’);Reject any references to ‘time to exhaustion’. 1

[5]

6.

(a) As the temperature falls so the diameter of the cluster decreases but clusterdiameter remaining constant at low temperature;the smaller the cluster the lower the surface area;will lose less heat / trap more heat; max. 2

(b) Axes correctly labelled with cluster temperature on x-axis;appropriate scale (allow if scale permits accurate plotting of all points);accuracy of plotting;curve of best fit or points joined with straight lines; 4

(c) (i) Increase in temperature produces increase in rate of enzymecontrolled reactions;link between respiration and enzymes;sugars for respiration [Award once];


(ii) The colder the cluster temperature, the more heat needs to beproduced;heat production coming from respiration;sugars for respiration [Award once]; max. 4

(d) Carbon dioxide production should follow a similar pattern to sugarconsumption / pattern described;this is an end product of respiration/also linked to rate of respiration; 2

(e) Fanning creates a current of air;evaporation of water;evaporation requires heat energy to vaporise water/reference to latent heat; 3

(f) (i) Acts as an insulator / trapping air;keeping body temperature high enough for activity;when insect leaves warmer colony / as cold at high altitude; max. 2

(ii) Would have been variation in hair length / long-haired andshort-haired / use of comparative;bees/colonies with longer haired bees bred / reproduced;passing on alleles/genes for longer hair to next generation; max. 3

[20]

7.

(a) Around day 2 / day 20;Coincides with LH / FSH peak; 2

(b) (No) / (no mark for no)Progesterone levels fail; 1

[3]

8.

(a) (i) (Few species adapted to) hostile/harsh environment/colonising bare/volcanic/fresh rock/little/no soil/water/humus; 1

(ii) Colonisers erode rock/release minerals;add (more) humus/increase nitrate content/organic material;when colonisers die/decompose; max. 2

(b) More/less competition; 1[4]

9.

(i) 4.25 – 3.35 = 0.9 tonnes ha–1

0.9/3.35 × 100%;= 26.9%; 2(Allow one mark for correct working)

(ii) Yield does not increase very much;Unlikely that value of crop will be greater than additional cost of fertiliser; 2

(iii) Fast acting;Components known;Concentrated therefore easy to store / apply;High nutrient content / value for money;


Can be applied evenly to a field;Can be applied using light machinery, reducing soil compaction; max 2

[6]

10l.

(a) Greater effect on straw / no significant difference/2% more on straw;Grain yield increases 30.1% / 30%;Straw yield increases 32.24% / 32%; 3

(b) (i) 324 / 622;= 0.524;

OR

622 / 324;= 1.92;One mark for working, one mark for correct answer 2

(ii) Grain is useful but straw is not / less useful; 1

(c) Competition for named factor; 2 max

Early LaterMore available forcrop / cropestablishes / later inseason, outcompeteweeds;

Less available forcrop / crop doesn’testablish well / whenweeds removedcannot make up lostproduction;

[8]

11.

(a) Energy losses due to radiation / evaporation / transpiration /in photosynthesis / energy of wrong wavelength / some ofenergy is heat;Extras: cancel 1

(b) 2920; 1

(b) (Ammonium) nitrite;Nitrite nitrate;

OR

Ammonium nitrate; (1 mark only)

If symbols: correct symbolse.g. ammonium ( nitrate (NO3) = NO MARKS

By nitrifying bacteria / Nitrosomonas / Nitrobacter / nitrification;By oxidation / using oxygen / aerobic; 3 max

[5]

12.

(a) (i) Cells of islets of Langerhans / pancreas / cells; 1


(ii) High glucagon because –eg stimulated by low glucose/glycogen must be converted to glucose;Low glucagon because –e.g. body has absorbed glucose; 2

(b) Clear understanding that a change from the set level initiates mechanismreturning it;low glucose secretion of glucagon glucose level increases/glucagon increase glucose increase glucagon decrease; 2

[5]

1.


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