(5) (1) (total 6 marks) - the student room

Q1. The Amazonian forest today contains a very high diversity of bird species.

• Over the last 2 000 000 years, long periods of dry climate caused this forest to separate into a number of smaller forests.

• Different plant communities developed in each of these smaller forests.

• Each time the climate became wetter again, the smaller forests grew in size and merged to reform the Amazonian forest.

(a) Use the information provided to explain how a very high diversity of bird species has developed in the Amazonian forest.

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(b) Speciation is far less frequent in the reformed Amazonian forest. Suggest one reason for this.

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(Total 6 marks)

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Q2. Sea otters were close to extinction at the start of the 20th century. Following a ban on hunting sea otters, the sizes of their populations began to increase. Scientists studied the frequencies of two alleles of a gene in one population of sea otters. The dominant allele, T, codes for an enzyme. The other allele, t, is recessive and does not produce a functional enzyme.

In a population of sea otters, the allele frequency for the recessive allele, t, was found to be 0.2.

(a) (i) Use the Hardy-Weinberg equation to calculate the percentage of homozygous recessive sea otters in this population. Show your working.

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

(ii) What does the Hardy-Weinberg principle predict about the frequency of the t allele after another 10 generations?

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(b) Several years later, scientists repeated their study on this population. They found that the frequency of the recessive allele had decreased.

(i) A statistical test showed that the difference between the two frequencies of the t allele was significant at the P = 0.05 level.

Use the terms probability and chance to help explain what this means.

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(ii) What type of natural selection appears to have occurred in this population of sea otters? Explain how this type of selection led to a decrease in the frequency of the recessive allele.

Type of selection ................................................................................

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

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(Total 7 marks)

Q3. Biologists studied the process of succession in an area of wasteland over a period of ten years. They calculated the index of diversity of the area every year. After three years, the index of diversity was 1.6. After ten years, it had risen to 4.3.

(a) What information concerning the organisms present in the area is suggested by the increase in the index of diversity?

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(b) The increase in the index of diversity is one indication that a biological succession is taking place in the area. Describe those features of a succession that would bring about an increase in the index of diversity.

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(Total 5 marks)

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Q4.S Red-green colour blindness is caused by a mutation in the gene coding for one of the opsin proteins which are needed for colour vision. The diagram shows the inheritance of red-green colour blindness in one family.

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Person 12 is pregnant with her fourth child. What is the probability that this child will be a male with red-green colour blindness? Explain your answer by drawing a genetic diagram. Use the following symbols

XR = an X chromosome carrying an allele for normal colour vision

Xr= an X chromosome carrying an allele for red-green colour blindness

Y = a Y chromosome

Probability = ...................................... (Total 4 marks)

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Q5. The diagram represents some of the light-independent reactions of photosynthesis.

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(a) Describe the light-independent reactions of photosynthesis and explain how they allow the continued synthesis of hexose sugars.

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(b) Describe the role of electron transport chains in the light-dependent reactions of photosynthesis.

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(c) Explain why the increase in the dry mass of a plant over twelve months is less than the mass of hexose produced over the same period.

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(Total 15 marks)

Q6. Ions of metals such as zinc often pollute rivers. The effect of zinc ions on gas exchange and respiration in fish was investigated. Fish were kept in tanks of water in a laboratory.

The fish in one group (X) had a solution of a zinc compound injected directly into their blood and were then put in a tank of zinc-free water. A second group (Y) was not injected but had the solution of the zinc compound added to the water in the tank.

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The partial pressure of oxygen in the blood of both groups of fish was then monitored. The results are shown in the graph.

(a) During this investigation, the water temperature in the tanks was kept constant. Explain why changes in the water temperature might lead to the results of the investigation being unreliable.

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(b) The results from the two groups were compared using a statistical test.

(i) Suggest a null hypothesis that could be tested.

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(ii) Explain why it is important to use a statistical test in analysing the results of this investigation.

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(c) Two suggestions were made to explain the results shown in the graph.

A Zinc ions reduce the rate at which oxygen is taken up from the water and passes into the blood.

B Zinc ions reduce the ability of haemoglobin to transport oxygen.

Which of these suggestions is the more likely? Explain the evidence from the graph that supports your answer.

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(d) During the investigation, the pH of the blood was also monitored. It decreased in group Y. Suggest an explanation for this decrease in pH.

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(e) Leaves were collected from sycamore trees growing in a polluted wood and the concentration of some metal ions in samples of these leaves was measured. Woodlice were then fed with the leaves. After 20 weeks, the concentration of the ions in the bodies of the woodlice was measured. Some of the results are shown in the table.

(i) Which of the elements shown in the table is concentrated most by the woodlice? Use suitable calculations to support your answer.

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Concentration of ions / µg g–1

Copper Cadmium Zinc Lead

Leaves 52 26 1430 908

Woodlice 1130 525 1370 132

(ii) Suggest what happens to most of the lead ions in the leaves eaten by the woodlice.

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(iii) Explain the difference in the copper ion concentration between the leaves and the woodlice.

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(f) Yorkshire fog is a species of grass. Two varieties of Yorkshire fog were studied. One variety was tolerant to arsenic, while the other variety was not. In a series of investigations, it was found that

• Arsenic-tolerant plants grow in soil which contains a high concentration of arsenic.

• Arsenic-tolerant plants growing in soil containing high concentrations of arsenic and phosphorus-containing compounds have very low concentrations of arsenic in their cells. They also have low concentrations of phosphates in their cells. Arsenic and phosphorus are chemically similar.

• Plants that are not tolerant to arsenic grow poorly on soil which has a high concentration of both arsenic and phosphorus-containing compounds.

• Tolerance to arsenic in Yorkshire fog is caused by a single gene with the allele, a, for tolerance recessive to the allele, A, for non-tolerance.

(i) What caused the allele for tolerance to first arise?

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(ii) Give two functions of phosphates in plant cells.

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(iii) Arsenic-tolerant Yorkshire fog plants are very rare in areas with low concentrations of arsenic in the soil, even where the soil has a high concentration of phosphate. Explain why they are unable to compete in these conditions with plants that are not tolerant to arsenic.

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(Total 20 marks)

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Q7. Chickens have a structure on their heads called a comb. The diagram shows four types of comb: walnut, pea, rose and single.

Two genes control the type of comb; each gene has a dominant and a recessive allele. The two genes are inherited independently, but interact to produce the four types of comb.

Genotype Phenotype

The symbol - indicates that either the dominant allele or recessive allele could be present

A- B- Walnut

A- bb Pea

aa B- Rose

aa bb Single

(a) A male with a pea comb, heterozygous for gene A, was crossed with a rose-combed female, heterozygous for gene B. Complete the genetic diagram to show the offspring expected from this cross.

Phenotypes of parents Pea comb Rose comb

Genotypes of parents ....................... .......................

Gametes formed ....................... .......................

Offspring genotypes .................................................................

Ratio of offspring phenotypes .................................................................

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(b) Chickens with rose or single combs made up 36% of one population. Assuming the conditions of the Hardy-Weinberg equilibrium apply, calculate the frequency of allele a in this population. Show how you arrived at your answer.

Frequency of allele a = ...................................... (2)

(Total 5 marks)

Q8. Tigers inhabit forests where they feed mainly on large prey animals. Over the past fifty years, there has been extensive deforestation in many areas where tigers are found.

The graph shows the relationship between the prey biomass of an area and the tiger population that the area can support.

(i) What is meant by the ecological term population?

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(ii) Use the graph to explain how deforestation might cause a reduction in the number of tigers in an area.

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Q9. Roundabouts are common at road junctions in towns and cities. Ecologists investigated the species of plants and animals found on roundabouts in a small town.

(a) Ground beetles are large black insects. The mark-release-recapture method can be used to estimate the ground beetle population on a roundabout. Describe how.

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(b) The grass on the roundabouts was mown at different time intervals. The table shows the mean number of plant species found on the roundabouts.

Mowing was also found to affect the number of insect species found on a roundabout. Use your knowledge of succession to explain how.

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Approximate interval between

mowing/days

Mean number of plant species

7 15.8

14 21.2

40 30.6

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(c) The carbon dioxide concentration was monitored at ground level in the centre of a small roundabout. The measurements were made on a summer day. Describe and explain how you would expect the concentration of carbon dioxide to fluctuate over the period of 24 hours.

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(Total 15 marks)

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Q10. People with night blindness have difficulty seeing in dim light. The allele for night blindness, N, is dominant to the allele for normal vision, n. These alleles are not carried on the sex chromosomes.

The diagram shows part of a family tree showing the inheritance of night blindness

(a) Individual 12 is a boy. What is his phenotype?

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(b) What is the genotype of individual 1? Explain the evidence for your answer.

Genotype .....................................................................................................

Evidence.......................................................................................................

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(c) What is the probability that the next child born to individuals 10 and 11 will be a girl with night blindness? Show your working.

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

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Q11. Deforestation often involves clearing large areas of forest for use as agricultural land.

(a) Deforestation reduces the diversity index of an area cleared in this way. Explain why.

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(b) Because the forest soil is often nutrient-poor, nitrogen-containing fertilisers may be applied to ensure good crop yields. Use your knowledge of the nitrogen cycle to explain the potential benefit of applying a fertiliser containing ammonium nitrate rather than one containing potassium nitrate.

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Q12. In fruit flies, the allele for grey body, G, is dominant to the allele for ebony body, g, and the allele for normal wings, N, is dominant to the allele for vestigial wings, n. Vestigial-winged flies, heterozygous for grey body colour, were crossed with ebony-bodied flies, heterozygous for normal wings.

Complete the genetic diagram to show the genotypes and phenotypes in this cross.

Parental phenotypes Grey body, vestigial wings Ebony body, normal wings

Parental genotypes .............................. ...............................

Gamete genotypes .............................. ...............................

Offspring genotypes ..............................................................................................

Offspring phenotypes ............................................................................................ (Total 4 marks)

Q13. (a) Explain how large-scale deforestation for agriculture would lead to a decrease in the diversity of organisms in the area.

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(b) Explain how large-scale deforestation could

(i) increase the concentration of carbon dioxide in the atmosphere in the area;

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(ii) decrease the concentration of carbon dioxide in the atmosphere in the area.

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(Total 5 marks)

Q14. (a) Explain the meaning of these ecological terms.

Population ....................................................................................................

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Community ...................................................................................................

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(b) Some students used the mark-release-recapture technique to estimate the size of a population of woodlice. They collected 77 woodlice and marked them before releasing them back into the same area. Later they collected 96 woodlice, 11 of which were marked.

(i) Give two conditions necessary for results from mark-release-recapture investigations to be valid.

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(ii) Calculate the number of woodlice in the area under investigation. Show your working.

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

(c) Explain how you would use a quadrat to estimate the number of dandelion plants in a field measuring 100 m by 150 m.

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(d) Two similar species of birds (species A and species B) feed on slightly different sized insects and have slightly different temperature preferences. The diagram represents the response of each species to these factors.

(i) Which of the numbered boxes describes conditions which represent

the niche of species A; .............

the niche of species B; .............

insects too small for species B and temperature too warm for species A; .............

insects too large for species A and temperature too cool for species B? ............. (2)

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(ii) These two species are thought to have evolved as a result of sympatric speciation. Suggest how this might have occurred.

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(Total 15 marks)

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Q15. (a) The flow chart shows the main stages in aerobic respiration.

(i) Complete the flow chart by writing, in the appropriate boxes, the number of carbon atoms in substance P and the name of substance Q.

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(ii) Some ATP is formed in the cytoplasm and some in the mitochondria. Use the information given to calculate the number of molecules of ATP formed in a mitochondrion from one molecule of glucose in aerobic respiration. Show how you arrived at your answer.

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(iii) In the presence of oxygen, respiration yields more ATP per molecule of glucose than it does in the absence of oxygen. Explain why.

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(b) Anabaena is a prokaryote found inside the leaves of a small fern. Anabaena can produce ammonia from nitrogen (nitrogen fixation). This reaction only takes place in the anaerobic conditions found in cells called heterocysts. Heterocysts are thick-walled cells that do not contain chlorophyll. The drawing shows the relationship between Anabaena and the fern.

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(i) Suggest how the features of the heterocysts improve the efficiency of the process of nitrogen fixation.

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(ii) In China, the fern is cultivated and ploughed into fields to act as an organic fertiliser. Explain how ploughing the fern plants into the soil results in an improvement in the growth of the rice crop grown in these fields.

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(Total 15 marks)

Q16. Finches are small birds. Fourteen species of finch are found on the Galapagos Islands.

(a) What is a species?

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(b) Measurements were made of the beak depth of two species of finch (species A and species B) on different islands. Species A is found on island 1, species B is found on island 2. Both species are found on island 3. They are thought to have colonised island 3 from islands 1 and 2 respectively. The graphs show the ranges of beak depths of the two species on the different islands.

What type of natural selection took place in the populations of both species after they had colonised island 3? Explain your answer.

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Q17. When coal is mined by open-cast mining, the top layer of soil is first scraped off and stored in a large heap. Once mining has finished, the area can be reclaimed. Soil from this store is then spread back over the surface.

Some of the bacteria living in the soil store respire aerobically and some respire anaerobically. Table 1 shows the numbers of aerobic and anaerobic bacteria found at different depths in a soil store.

Table 1

(a) Some of the soil used to determine bacterial numbers was collected from the surface of the soil store. Describe how you would ensure that this soil was collected at random.

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Depth / cm

Mean number of bacteria per gram of soil (× 107)

Aerobic bacteria Anaerobic bacteria

after 1 month after 6 months after 1 month after 6 months

0 12.0 12.1 0.6 0.8

50 10.4 8.6 0.8 1.3

100 10.1 6.1 0.7 4.1

150 10.0 3.2 0.7 7.9

200 11.6 0.8 0.7 8.4

250 11.9 0.7 0.8 8.8

300 11.0 0.8 0.6 9.1

(b) (i) Describe how the numbers of aerobic bacteria after 6 months change with depth.

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(ii) Explain the difference in the numbers of aerobic bacteria at a depth of 300 cm between 1 and 6 months.

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(c) Explain how the changes in bacterial numbers which take place at 150 cm illustrate the process of succession.

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Dehydrogenase is an enzyme involved in aerobic respiration. Dehydrogenase activity in a soil sample can be used as a measure of the activity of aerobic bacteria. The graph shows the mean dehydrogenase activity of soil samples taken from the same depth in a soil store at different times. The bars on the graph represent two standard errors above and below the mean.

(d) (i) From what depth in the soil store would you expect these soil samples to have been taken? Use information from Table 1 to explain your answer.

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(ii) How would you expect dehydrogenase activity to vary with depth after 6 months?

Use information from Table 1 to explain your answer.

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(e) What do the error bars tell you about the difference between the mean dehydrogenase activity at 6 months and 3 years? Explain your answer in terms of probability and chance.

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(f) Table 2 shows the dehydrogenase activity and the number of aerobic bacteria present in some soil samples.

Table 2

A sample of soil was found to have dehydrogenase activity of 8.7 arbitrary units. Explain how you would use the data in Table 2 to predict the likely number of aerobic bacteria in 1 g of this soil sample.

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(Total 20 marks)

Dehydrogenase activity / arbitrary units

Number of aerobic bacteria per gram of soil (× 107)

13.1 12.0

9.2 8.7

5.5 6.5

3.0 4.6

2.2 2.7

0.4 0.6

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Q18. (a) Fertilisers are added to soils to replace the nutrients lost when crops are harvested.

Give two advantages of using

(i) an organic fertiliser such as farmyard manure;

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(ii) an inorganic fertiliser.

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(b) The table shows the effects of adding manure or inorganic fertiliser to some crops grown in plots.

(i) How should the control plot be treated?

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Yield of crop / tonnes per hectare

Crop Control plot Farmyard manure only

Inorganic fertiliser only

Sugar beet 3.8 15.6 15.6

Mangold 3.8 22.3 30.9

Wheat 2.1 3.5 3.1

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(ii) Suggest why inorganic fertiliser improved the yield of the mangold crop more than the sugar beet crop.

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(Total 7 marks)

Q19. (a) The table contains some statements relating to biochemical processes in a plant cell. Complete the table with a tick if the statement is true or a cross if it is not true for each biochemical process.

(4)

Statement Glycolysis Krebs cycle Light-dependent reaction of

photosynthesis

NAD is reduced

NADP is reduced

ATP is produced

ATP is required

(b) An investigation was carried out into the production of ATP by mitochondria. ADP, phosphate, excess substrate and oxygen were added to a suspension of isolated mitochondria.

(i) Suggest the substrate used for this investigation.

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(ii) Explain why the concentration of oxygen and amount of ADP fell during the investigation.

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(iii) A further investigation was carried out into the effect of three inhibitors, A, B and C, on the electron transport chain in these mitochondria. In each of three experiments, a different inhibitor was added. The table shows the state of the electron carriers, W–Z, after the addition of inhibitor.

Give the order of the electron carriers in this electron transport chain. Explain your answer.

Order .............. .............. .............. ..............

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

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(Total 9 marks)

Inhibitor added

Electron carrier

W X Y Z

A oxidised reduced reduced oxidised

B oxidised oxidised reduced oxidised

C reduced reduced reduced oxidised

Q20. In the activated sludge method of sewage treatment, organic matter in untreated sewage supplies nutrients to bacteria in the treatment tank. These bacteria include decomposers and nitrifying bacteria. The bacteria are eaten by ciliated protoctistans, which are, in turn, eaten by carnivorous protoctistans.

(a) (i) Sketch and label a pyramid of energy for the organisms found in the treatment tank.

(1)

(ii) Explain what causes this pyramid of energy to be this shape.

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(b) (i) Explain the roles of the decomposers and the nitrifying bacteria in converting nitrogen in organic compounds in the sewage into a soluble, inorganic form.

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(ii) Nitrifying bacteria are one kind of bacteria that are important in the nitrogen cycle; nitrogen-fixing bacteria are another kind. Describe the part played by nitrogen-fixing bacteria in the nitrogen cycle.

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(c) The organic matter in untreated sewage consists of small particles, which are suspended in water. Activated sludge consists of solid lumps (flocs) of organic matter and bacteria. When the two are mixed in the treatment tank, bacteria from the flocs become dispersed in the water and feed on the suspended organic matter, converting it to flocs. Different types of ciliated protoctistans feed on the bacteria.

• Free-swimming protoctistans are able to move throughout the tank.

• Crawling protoctistans can only move over the surface of the flocs.

The diagram shows the change in the nature of the organic matter in the treatment tank and the changes in the numbers of the different types of organisms present.

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(i) Explain the changes in the numbers of dispersed bacteria and the numbers of free-swimming protoctistans.

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(ii) Explain how the changes that occur in the treatment tank illustrate the process of succession.

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(Total 15 marks)

Q21. (a) Name the type of bacteria which convert

(i) nitrogen in the air into ammonium compounds;

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(ii) nitrites into nitrates.

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(b) (i) Other than spreading fertilisers, describe and explain how one farming practice results in addition of nitrogen-containing compounds to a field.

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(ii) Describe and explain how one farming practice results in the removal of nitrogen-containing compounds from a field.

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(Total 6 marks)

Q22. Purification ponds can be used in warm climates to break down sewage. The ponds are about 1m deep and contain bacteria and green algae. The diagram summarises the processes involved in the breakdown of sewage in a purification pond.

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(a) Explain the advantage of having both algae and bacteria in a purification pond.

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S (b) Purification ponds only work efficiently when they are shallow and warm. Explain why.

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(Total 8 marks)

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Q23. Duchenne muscular dystrophy is a sex-linked inherited condition which causes degeneration of muscle tissue. It is caused by a recessive allele. The diagram shows the inheritance of muscular dystrophy in one family.

(a) Give evidence from the diagram which suggests that muscular dystrophy is

(i) sex-linked; ...........................................................................................

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

(ii) caused by a recessive allele. ...............................................................

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

(b) Using the following symbols,

XD = an X chromosome carrying the normal allele

Xd = an X chromosome carrying the allele for muscular dystrophy

Y = a Y chromosome

give all the possible genotypes of each of the following persons.

5 ..................................................................................................................

6 ..................................................................................................................

7 ..................................................................................................................

8 .................................................................................................................. (2)

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(c) A blood test shows that person 14 is a carrier of muscular dystrophy. Person 15 has recently married person 14 but as yet they have had no children. What is the probability that their first child will be a male who develops muscular dystrophy?

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

(Total 5 marks)

Q24. Great tits are small birds. The graph shows the relationship between the number of breeding pairs in the population and the mean number of eggs per nest in different years in a wood.

(a) Explain the relationship shown by the graph.

......................................................................................................................

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

S (b) Female great tits usually lay between 3 and 14 eggs in a nest.

(i) In the same year, the birds do not all lay the same number of eggs. Explain how one factor, other than the number of breeding pairs, could influence the number of eggs laid by a great tit.

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(ii) Natural selection influences the number of eggs laid. Explain why great tits that lay fewer than 3 eggs per nest or more than 14 eggs per nest are at a selective disadvantage.

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(Total 6 marks)

Q25. In a breed of cattle the H allele for the hornless condition is dominant to the h allele for the

horned condition. In the same breed of cattle the two alleles CR (red) and CW (white) control coat colour. When red cattle were crossed with white cattle all the offspring were roan. Roan cattle have a mixture of red and white hairs.

(a) Explain what is meant by a dominant allele.

......................................................................................................................

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

(b) Name the relationship between the two alleles that control coat colour.

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(c) Horned, roan cattle were crossed with white cattle heterozygous for the hornless condition. Compete the genetic diagram to show the ratio of offspring phenotypes you would expect.

Parental phenotypes Horned, roan × hornless, white

Parental genotypes

Gametes

Offspring genotypes

Offspring phenotypes

Ratio of offspring phenotypes

(4)

(d) The semen of prize dairy bulls may be collected for in vitro fertilisation. The sperms in the semen can be separated so that all the calves produced are of the same sex. The two kinds of sperms differ by about 3% in DNA content.

(i) Explain what causes the sperms of one kind to have 3% more DNA than sperms of the other kind.

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

(ii) Suggest one reason why farmers would want the calves to be all of the same sex.

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

(Total 9 marks)

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Q26. The wildebeest is a large mammal that lives on grasslands in Africa and feeds on a number of species of plant. A lot of rain falls from April to May and also in November. In the dry season between July and October very little rain falls.

The graph shows changes in the mean protein content of all the plants that could be eaten at different times of year. It also shows the mean protein content of the food the wildebeest actually eat.

S (a) During the dry season the protein content of the plants decreases. Suggest one way in which a lack of rain could account for this change.

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

S (b) Throughout the year the mean protein content of all the plants which could be eaten and the mean protein content of the food actually eaten differs. Suggest one explanation for this difference.

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S (c) When wildebeest eat food containing less than 6% protein, they start to lose protein from their body tissues. Suggest and explain how a deficiency of one named protein makes the wildebeest more susceptible to being caught by predators.

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

(Total 6 marks)

Q27. (a) Explain what is meant by

(i) succession;

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

(ii) a climax community.

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Heather plants are small shrubs. Heather plants are the dominant species in the climax community of some moorlands. The structure and shape of a heather plant changes as it ages. This results in changes in the species composition of the community. A large area of moorland was burnt leaving bare ground. The table shows four stages of succession in this area.

Time after burning / years

Appearance of heather plant

Mean percentage

cover of heather

Other plant species present

4

10 Many

12

90 Few

19

75 Several

24

30 Many

(b) Explain why the number of other plant species decreases between 4 and 12 years after burning.

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

S (c) The rate at which a heather plant produced new biomass was measured in g per kg of heather plant per year. This rate decreased as the plant aged. Use the information in the table to explain why.

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(Total 8 marks)

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Q28. The wildebeest is a large mammal that lives on grasslands in Africa and feeds on a number of species of plant. A lot of rain falls from April to May and also in November. In the dry season between July and October very little rain falls.

The graph shows changes in the mean protein content of all the plants that could be eaten at different times of year. It also shows the mean protein content of the food the wildebeest actually eat.

S (a) During the dry season the protein content of the plants decreases. Suggest one way in which a lack of rain could account for this change.

......................................................................................................................

......................................................................................................................

......................................................................................................................

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

S (b) Throughout the year the mean protein content of all the plants which could be eaten and the mean protein content of the food actually eaten differs. Suggest one explanation for this difference.

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S (c) When wildebeest eat food containing less than 6% protein, they start to lose protein from their body tissues. Suggest and explain how a deficiency of one named protein makes the wildebeest more susceptible to being caught by predators.

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

(Total 6 marks)

Q29. (a) Explain how including leguminous plants in a crop rotation reduces the need to use artificial fertilisers.

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

(b) The graph shows the effects of applying potassium fertiliser at different rates to a crop of wheat.

Explain how the graph shows the law of diminishing returns.

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S (c) Application of very high concentrations of fertiliser to the soil causes plants to wilt. Explain why.

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

(Total 6 marks)

Q30. (a) Explain how the use of insecticides may poison the animals at the top of a food chain.

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

The diagram shows a hedgerow and part of a field with a crop. The land is farmed in a way that conserves wildlife. The strip of bare ground next to the hedgerow is ploughed frequently to prevent any plants from growing. The first 6 m of the field, called the conservation headland, is sprayed with a selective herbicide to control some kinds of weeds. The rest of the field is sprayed with herbicide to kill all weeds.

Hedgerow Bare Conservation Crop ground headland

(2 m wide) (6 m wide)

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S (b) Suggest one advantage of leaving a strip of bare ground between the hedgerow and the field.

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

(c) Suggest the benefit of allowing some weeds to grow in the conservation headland.

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

S (d) After harvesting the crop, the farmer digs the unwanted stems and roots into the soil. Explain how the nutrients contained in these plant parts become available for use by other organisms.

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(Total 9 marks)

Q31. (a) In a demographic transition, give one factor that might cause

(i) an increase in the birth rate; ...............................................................

.............................................................................................................

(ii) a decrease in the death rate. ..............................................................

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

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(b) Births, deaths and migration affect population growth. The graph shows the effects of these factors on a human population between 1960 and 2000. During this period the death rate was almost constant.

(i) From the information given, what does the graph show about changes in birth rate between 1960 and 1980? Explain your answer.

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

(ii) Describe the effect of immigration and emigration on the growth of this population between 1960 and 2000.

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

(Total 6 marks)

Q32. S Clover plants have leaves all through the year. Some clover plants have leaves that produce poisonous hydrogen cyanide gas when damaged. These cyanogenic plants are less likely to be eaten by snails. However, the leaves of these plants can be damaged by frost, resulting in the production of enough hydrogen cyanide to kill the plants. Acyanogenic plants do not produce hydrogen cyanide. This characteristic is genetically controlled.

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The map shows the proportions of the two types of plant in populations of clover from different areas in Europe. It also shows isotherms, lines joining places with the same mean January temperature.

(a) Explain how different proportions of cyanogenic plants may have evolved in populations in different parts of Europe.

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(b) Differences in cyanide production may affect the total number of clover plants growing in different areas. Describe how you would use quadrats in an investigation to determine whether or not there is a difference in the number of clover plants in two large areas of equal size.

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(Total 8 marks)

Q33. Essay

You should write your essay in continuous prose.

Your essay will be marked for its scientific accuracy.

It will also be marked for your selection of relevant material from different parts of the specification and for the quality of your written communication.

The maximum number of marks that can be awarded is

Write an essay on the following topic:

The transfer of substances containing carbon between organisms and between organisms and the environment.

(Total 25 marks)

Scientific Breadth of knowledge Relevance Quality of written communication

16 3 3 3

Q34. (a) Name the two substances produced by anaerobic respiration in humans.

1 ...................................................................................................................

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

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(b) When an athlete runs in a 100 metre race, 90% of the energy needed is provided by anaerobic respiration.

(i) Explain why most of the energy is provided by anaerobic respiration rather than aerobic respiration.

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.............................................................................................................

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

(ii) The athlete continues to breathe deeply for several minutes after the race ends. Explain why this is necessary.

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

(Total 6 marks)

Q35. (a) Name the two substances produced by anaerobic respiration in humans.

1 ...................................................................................................................

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

(b) When an athlete runs in a 100 metre race, 90% of the energy needed is provided by anaerobic respiration.

(i) Explain why most of the energy is provided by anaerobic respiration rather than aerobic respiration.

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(ii) The athlete continues to breathe deeply for several minutes after the race ends. Explain why this is necessary.

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

(Total 6 marks)

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M1. (a) 1. No interbreeding / gene pools are separate / geographic(al) isolation;

Accept: all marks if answer written in context of producing increased diversity of plants 1 Do not award this mark in context of new species being formed and then not interbreeding 1 Accept reproductive isolation as an alternative to no interbreeding

2. Mutation; 2 Accept: genetic variation

3. Different selection pressures / different foods / niches / habitats; 3 Accept: different environment / biotic / abiotic conditions or named condition 3 Neutral: different climates

4. Adapted organisms survive and breed / differential reproductive success;

5. Change / increase in allele frequency / frequencies; 5

(b) Similar / same environmental / abiotic / biotic factors / similar / same selection pressures / no isolation / gene flow can occur (within a species);

Accept: same environment 1

[6]

M2. (a) (i) Two marks for correct answer of 4;;

One mark for calculation involving 0.2 × 0.2 or 0.04; 2

(ii) 0.2/ the frequency remains the same; Reject if wrong frequency is quoted

1

(b) (i) 1. There is a probability of 5%/0.05;

2. That difference in frequencies / difference in results are due to chance; Accept 95% probability changes in frequencies not different as a result of chance

2

(ii) 1. Directional;

2. The recessive allele confers disadvantage/ the dominant allele confers advantage/more likely to survive / reproduce;

Assume "it" to refer to the recessive allele 2. References to selection do not gain credit as the term is in the question. Allow reference to phenotype / enzyme functionality (instead of allele) when describing advantage/disadvantage.

2 [7]

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M3. (a) Increase in number of species;

Increase in numbers of some species; 2

(b) Initial environment hostile / few organisms adapted;

These organisms change the environment / suitable example;

More niches / more habitats;

Allowing other organisms to become established; max. 3

[5]

M4. parental genotypes correct: X X AND X Y;

gametes correct for candidate’s parental genotypes; offspring genotypes correct and colourblind male identified as X Y /

correct genotypes derived from cand’s gametes and identify X Y;

correct probability = ¼ / 0.25 / 25% / 1 in 4 / 1:3 ; [4]

M5. (a) 1 5C/RuBP combines with CO2;

2 to form 3C compound / TP / GP;

3 using ATP;

4 and reduced NADP / eq;

5 2 molecules of 3C compound/ TP / GP form hexose;

6 all RuBP is regenerated;

7 10 molecules of 3C/TP/GP form 6 molecules of 5C/RuBP; 6 max

R r R

r

r

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(b) 1 electron transport chain accepts excited electrons;

2 from chlorophyll / photosystem;

3 electrons lose energy along chain;

4 ATP produced;

5 from ADP and Pi;

6 reduced NADP formed;

7 when electrons (from transport chain) and H combine with NADP;

8 H from photolysis;

6 max

(c) 1 some hexose/biomass/eq. used in respiration; growth cancels this point

2 CO2 produced (is lost to air);

3 some parts of the plant are eaten;

4 some parts lost to decomposers / in leaf fall; 3 max

[15]

M6. (a) (variation in) temperature will affect the solubility of oxygen/ rate of respiration / use of oxygen by cells/ diffusion/ gas exchange; to gain credit point made must concern oxygen

1

(b) (i) there is no difference between the partial pressure of oxygen in the two groups / the partial pressure of oxygen is the same in each group;

1

(ii) results may have been due to chance; statistical test allows us to determine the probability of this / of the difference between results being significant; enables acceptance or rejection of null hypothesis; The key points here are chance and probability used in the correct context.

2 max

(c) A; because partial pressure of oxygen only reduced when zinc in water / in Y / because when injected zinc / in X has no effect on partial pressure of oxygen in blood;

2

+

+

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(d) less oxygen transport to cells / in fish / in blood; anaerobic respiration; lactic acid produced / less carbon dioxide removed (from gills); more H ;

3 max

(e) (i) copper; calculation based on comparing concentration in woodlice with that in leaves; accept any suitable method here, giving marks for the method and explanation. For example, calculating ratio of concentration in woodlice to concentration in leaves.

2

(ii) not absorbed from gut / passes out in faeces/ egested / urine / excreted;

1

(iii) woodlice eat large amount of leaves; copper stored/accumulates in body;

2

(f) (i) mutation; 1

(ii) (as a component of) nucleic acids / DNA / RNA / nucleotides; phospholipids; ATP/ADP;

2 max

(iii) arsenic-tolerant plants would not be able to take up phosphates / take up a little phosphate; since likely to involve same mechanism/same carrier/protein; (process of ) growth would be poorer than non-tolerant plants;

3 [20]

M7. (a) Parents genotypes Aabb aaBb ;

Gametes formed Ab ab aB ab ; if parental genotypes wrong allow correctly derived gametes only

Offspring genotypes AaBb Aabb aaBb aabb

and

Offspring phenotypes 1 Walnut ; 1 Pea : 1 Rose : 1 single ; Just one mark for offspring genotypes and phenotypes If parents not diploid, no marks gained

3

+

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(b) Correct answer 0.6, however derived, scores 2 marks Wrong answer, but evidence of correct working

(e.g. p /q = 0.36) scores 1 mark 2

[5]

M8. (i) Population is the total number of organisms/individuals of a species/tigers in an area (at a given time);

1

(ii) (Deforestation involves) habitat destruction/ destruction of niches;

Some prey animals move out or die / fewer suitable prey for tiger/ less food for tiger; Reduces tiger population if prey biomass falls below 600 (tonnes per km );

3 [4]

M9. (a) 1 Sample of ground beetles captured and counted (a); 2 Released and second sample captured; 3 Count total number of beetles (B) and number marked (b);

4 Total population (A) estimated from the relationship ; 5 Detail of method e.g. pitfall trap/marking with tippex; 6 Refinement to ensure greater accuracy e.g. large number/

marking in position such that does not affect survival; 5 max

(b) 1 Mowing prevents growth of woody plants; 2 By cutting off growing point; 3 The longer the interval between mowing, the further succession

can progress; 4 With frequent mowing diversity of plants will be less; 5 Fewer insect inhabitants/niches available;

Q Since this is an ecological question, use of appropriate ecological terminology is expected. Credit such terms as producer, consumer, habitat, and niche. Do not credit inappropriate terminology such as “places” to live and “fighting for food”.

5

2 2

2

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(c) 1 Higher carbon dioxide concentration at night/during darkness; 2 Photosynthesis only takes place during light; 3 Photosynthesis removes carbon dioxide and respiration adds

carbon dioxide; 4 Respiration taking place throughout 24 hours; 5 Quantitative consideration such as that in plants overall

photosynthetic rate greater than respiration rate; 6 Human effect such as additional carbon dioxide from heavy

daytime traffic/street lighting could prolong photosynthesis; 5 max

[15]

M10. (a) Normal sight; 1

(b) Nn; Must have at least one N allele as she has the condition and must pass on an n allele to her normal sighted children;

2

(c) Two marks for correct answer of ¼ / 0.25 / 25%; One mark for incorrect answer that determines probability of next child having night blindness as ½ / 0.5 / 50%;

2 max [5]

M11. (a) deforestation removes many habitats/niches fewer species/ fewer types of organisms;

(do not credit just fewer organisms); 2

(b) 1. ammonium nitrate contains more nitrogen per molecule than potassium nitrate; 2. nitrate ions in fertiliser available/ absorbed immediately; 3. ammonium converted to nitrate; 4. by nitrifying bacteria/Nitrosomonas and Nitrobacter; 5. fertiliser would provide only the initial release of nitrate/ potassium nitrate;

3 max [5]

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M12. Parental genotypes: Gg nn gg Nn ; Gamete genotypes Gn gn gN gn ;

All offspring genotypes correct;

All offspring genotypes correctly derived; [4]

gN gn

Gn

Gg Nn

Grey, normal

Gg nn

Grey, vestigial

gn

gg Nn

Ebony, normal

gg nn

Ebony, vestigial

M13. (a) Removal of forest removes many ecological niches/habitats/food sources/shelter; Reduces numbers of species that can exist in the area;

2

(b) (i) Reduce amount of CO 2 used in photosynthesis;

increase amount of CO 2 produced in combustion/decomposition;

(ii) Less respiration; By plants/animals/decomposers;

max 3 [5]

M14. (a) Population – organisms of one species in an ecosystem/habitat/area; Community – organisms of all species / all populations in an ecosystem/habitat/ area;

2

(b) (i) No immigration/migration (Ignore references to emigration); No reproduction (Ignore references to death); Idea of mixing; Marking does not influence behaviour / increase vulnerability to predation; Sample/population large enough;

max 2

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(ii) ; 672;

Correct answer (however derived) scores 2 marks Incorrect answer with evidence of correct method scores 1 mark.

2

(c) Principle of randomly placed quadrats; Method of producing random quadrats; (Reject ‘throwing’) Valid method of obtaining no. dandelions in given area (mean per quadrat/ total no. in many quadrats); Multiply to give estimate for total field area;

max 3

(d) (i) Niche of A – 1; Niche of B – 3; Too small for B / too hot for A – 4; Too large for A / too cold for B – 2; All four correct = 2 marks; any 2 correct = 1 mark

2

(ii) Original population living in one area / 2 species evolved in the area; Idea of genetic variability; Concept of reproductive isolation; Possible mechanism; Gene pools become increasingly different; Until interbreeding does not produce fertile offspring;

max 4 [15]

M15. (a) (i) P = 3; Q = acetylcoenzyme A;

2

(ii) 36 ATP, however derived = 2 marks 30 ATP, however derived = 1 mark

2

(iii) Correct statement in the context of aerobic respiration or anaerobic respiration concerning: Oxygen as terminal hydrogen/electron acceptor; Operation of electron transport chain/ oxidative phosphorylation; Fate of pyruvate; Krebs cycle; Significance of ATP formed in glycolysis;

max. 3

(b) (i) Thick walls exclude oxygen; Produced by photosynthetic cells (of fern and Anabaena); Contain no chlorophyll so do not photosynthesise; Do not produce oxygen; Oxygen would inhibit nitrogen fixation process;

max. 3

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(ii) Decomposers/ bacteria/fungi/saprobionts (in fields);

Convert protein/organic nitrogen (in cells of fern) into ammonium ions (allow ammonia); Ammonium ions (ammonia) converted to nitrite; Nitrite converted to nitrate;

Allow 1 mark for NH3 /NH NO3

By nitrifying bacteria / correctly named; Nitrate used to form protein / amino acids in rice; Link between application of fern and protein/cells of rice; Decomposers respire (suitable substrate) and release CO

2;

Used in photosynthesis by rice; max. 5

[15]

M16. (a) group of organisms with similar features; can (interbreed to) produce fertile offspring;

2

(b) directional selection; any TWO from selection against one extreme / for one extreme; against broadest beaks in B and narrowest beaks in A / for narrowest in B and broadest in A; whole distribution / range / mean / mode / median is shifted towards favoured extreme;

3 max [5]

M17. (a) Tapes / string / axes laid out at right angles / grid area; Method of obtaining random co-ordinates; Do not allow “Use random number generator”

2

(b) (i) Decrease then remain constant; From 200 cm / over 150 cm;

2

(ii) Oxygen decreasing because soil becomes more compacted/ not replaced; Decrease in oxygen leads to fewer aerobes surviving; Respiration;

max 2

(c) Anaerobic bacteria replace aerobic; As oxygen decreased by aerobic bacteria; Remove competition; Aerobic bacteria no longer able to survive in these conditions;

max 3

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(d) (i) Near the surface / in top 50 cm; Table shows decrease with time at greater depths;

2

(ii) Decrease; Fewer aerobic bacteria with depth; Oxygen concentration decreases / less oxygen at depth;

3

(e) Probability greater than 95% / 0.95; Results are not due to chance / results are significant; Because bars do not overlap;

3

(f) Plot as graph; Draw line of best fit; Read off appropriate value;

3 [20]

M18. (a) (i) More micronutrients / greater range of nutrients; Nutrients released slowly; Improves soil quality / adds humus / adds microbes / improves soil structure; Improves water-holding capacity of soil / reduces leaching/eutrophication; Improves soil aeration; Already available;

max 2

(ii) Known nutrient content; Nutrients available immediately/fast acting; Nutrients distributed evenly; Doesn’t contain pests; Better to handle / easy to use / easy to store/transport; Concentrated in nutrients / needed in smaller amounts; Applied using light machinery so avoids soil compaction;

max 2

(b) (i) Same as other plots / named variable controlled; Without fertiliser;

2

(ii) Contains a nutrient/nutrients important for mangolds / Idea that different crops have different nutrient requirements / Inorganic fertiliser contains ingredient which inhibits beet growth;

max 1 [7]

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M19. (a) x;

x x ;

x x 4

(b) (i) pyruvate/succinate/any suitable Krebs cycle substrate; 1

(ii) ADP and phosphate forms ATP; oxygen used to form water / as the terminal acceptor;

2

(iii) Y X W Z; order of carriers linked to sequence of reduction / reduced carriers cannot pass on electrons when inhibited;

2 [9]

M20. (a) (i) pyramid correctly drawn and labelled; ignore organic matter

1

(ii) energy lost/not transferred between trophic levels; in respiration /as heat / in excretory products / movement;

ignore in urea / in faeces. ‘Growth’ cancels 2 marking point only 2

(b) (i) decomposers convert (nitrogen in organic compounds) into ammonia/ammonium; suitable example of “organic nitrogen” - protein/urea/amino acid etc. (e.g. linked to process); nitrifying bacteria / correctly named convert ammonium to nitrate; via nitrite;

3 max

(ii) convert nitrogen (gas) into ammonium / ammonia / amino acids; add usable/available nitrogen to an ecosystem / eq.;

2

(c) (i) 1 numbers of dispersed bacteria increase as they feed on organic matter;

2 numbers of free-swimming protoctistans increase because number of bacteria increase;

3 dispersed bacteria decrease as amount of dispersed organic matter decreases / due to lack of food / as organic matter is converted to flocs;

4 decrease as are preyed on by free-swimming protoctistans;

5 decrease in free-swimming protoctistans due to lack of dispersed bacteria;

3 max

nd

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(ii) 1 (in a succession) organisms (enter an area and) change the environment/conditions;

2 creating new niches / habitats;

3 allows different species / different types of organisms to enter / be successful;

4 dispersed bacteria change dispersed organic matter to flocs;

5 presence of flocs allows crawling protoctistans to enter / to increase / to be successful;

4 max [15]

M21. (a) (i) nitrogen-fixing;

(ii) nitrifying; (names neutral, name only no mark)

2

(b) (i) growing legumes/ named legume; ploughed in/allowed to decompose/nitrogen-fixing (bacteria in nodules);

OR

allow cattle/named species/(farm) animals (to graze); add dung/urine;

OR

spread/add manure/slurry; decomposed to release nitrates/ammonia/nitrites;

2

(ii) bare soil/fallow in winter/hedge removal; leaching (of nitrates)/soil erosion;

OR

uptake of nitrates/ammonium compounds by crop; harvesting crop/named crop which would be harvested;

OR

(farm) animals eat plants (in field); (then) animals removed;

2 [6]

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M22. (a) breakdown of organic matter/sewage by enzymes from bacteria; nitrates/ammonia used by algae to make amino acids/proteins; algae photosynthesise; bacterial respiration uses O

2 /produces CO

2 for algae;

(respiration) allows for reproduction/growth of bacteria; 4

(b) sufficient light penetration for photosynthesis (of algae); warm leads to faster enzyme activity; faster bacterial respiration/decomposition; faster photosynthesis; increased growth/reproduction of bacteria/algae;

4 [8]

M23. (a) (i) Only seen in males / not in females; 1

(ii) Unaffected parents/mother → child with M.D./ (1 ×)2 → 5 / (3 ×) 4 → 11 / 8 (× 9) → 13;

1

(b) 5 = XdY

6 = XDY

7 = XDXd AND XDXD

8 = XDXd;;

All 4 correct = 2 marks 2 or 3 correct = 1 mark

max 2

(c) ¼ / 0.25 / 25% / 1:3 / 1 in 4; (NOT ‘1:4’) 1

[5]

M24. (a) principle of intraspecific competition; for amount of food available; more energy needed to find food/less energy to produce eggs;

OR

number of territories; more energy spent fighting/defending territory;

OR

availability as prey; predators spend less time searching for nests;

2 max

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(b) (i) age of bird - young or old birds produce fewer eggs; time of breeding - early or late breeding less food available/ temperature effect; genotype - variation in genetic ability to produce eggs; quality of territory - description of some relevant resource in territory; (reject food as resource in territory if given in(a)) predation of eggs - lays more to replace eaten eggs;

1 max

(ii) when high number of eggs, each individual young will receive less food; reference to mortality rates to disease/predators for low numbers of eggs; so in both cases low number of offspring will reach maturity/survive; so less likely to pass on genes/alleles;

3 [6]

M25. (a) is always expressed(in the phenotype) / produces (functional) proteins; 1

(b) codominance; 1

(c) Parental geneotypes - hhC C , HhC C ;

Gametes- Offspring geneotypes - HhC C , hhC C , HhC C , hhC C ; Offspring pheneotypes - hornless horned hornless horned

roan roan white white Ratio of offspring - 1 1 1 1;

4

(d) (i) sperm(with more DNA) have X chromosome; X is larger / has more genes than Y;

2

(ii) female for milk / males for meat / male or female for breeding; 1

[9]

R w w w

R w R w w w w w

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M26. (a) less nitrate taken up; less amino acid/protein synthesis;

OR

parts of plant higher in protein die; higher proportion of cellulose/non-protein components in diet;

2

(b) (wildebeest) selective feeders/only some species/parts of plant eaten; choose to eat species/part of plant with high protein content;

2

(c) named protein; consequences of lack of protein related to failure to escape from predators;

examples:

myosin/actin; (skeletal) muscles weak/less muscular tissue so slower movement;

OR

relevant named enzyme; why deficiency of enzyme increases chance of being caught;

OR

haemoglobin; insufficient oxygen for muscle contraction;

2 [6]

M27. (a) (i) change in community over time; either due to change environmental/abiotic factors / change is due to species present;

2

(ii) stable community/no further succession/final community; 1

(b) (increased) interspecific competition; for light/nutrients/named nutrient/water;

2

(c) fewer leaves/lower surface area/shading of leaves; less photosynthesis to produce new biomass/glucose/growth; competition with other species for nitrates/named nutrient; reduced synthesis of protein or named compound; ratio of leaves to woody parts and roots decreases; so higher respiration relative to photosynthesis;

3 max [8]

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M28. (a) less nitrate taken up; less amino acid/protein synthesis;

OR

parts of plant higher in protein die; higher proportion of cellulose/non-protein components in diet;

2

(b) (wildebeest) selective feeders/only some species/parts of plant eaten; choose to eat species/part of plant with high protein content;

2

(c) named protein; consequences of lack of protein related to failure to escape from predators;

examples:

myosin/actin; (skeletal) muscles weak/less muscular tissue so slower movement;

OR

relevant named enzyme; why deficiency of enzyme increases chance of being caught;

OR

haemoglobin; insufficient oxygen for muscle contraction;

2 [6]

M29. (a) contain nitrogen-fixing bacteria in roots/nodules (so don’t need fertiliser); nitrogen containing compounds added to the soil when plant dies/after harvest of crop;

2

(b) increase in yield up to 500-600 kg ha ;

at 500-600 kg ha rate of increase slows/ no significant increase (with extra fertiliser);

2

(c) low(er)/more negative water potential in soil (than in the plant); prevents roots from taking up water (from the soil); plants still lose water by transpiration; plants lose water to soil by osmosis;

2 max [6]

–1

–1

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M30. (a) (accumulates) in (fatty) tissue/ is not excreted/ not metabolised/broken down; becomes concentrated higher up the food chain/ bioaccumulation/ biomagnification;

2

(b) prevents disease/pest organisms from reaching crop plants/prevents herbicides from reaching hedgerow/enables machinery to manoeuvre without damaging crop/hedgerow;

1

(c) some weeds provide habitats/niche for (beneficial) insects/animals: allow (insect) pest predators to survive; conserve (common) weed plants; weeds are producers in food chains/food source;

2 max

(d) decomposers/saprophyte/ bacteria/ fungi /micro organisms; (organisms) excrete/ produce nitrogenous waste/ e.g.; bacteria convert to nitrate/nitrifying bacteria; (increased) nitrates(in soil) taken up/used by plants; release of phosphate/potassium; organisms respire and produce carbon dioxide; used by plants in photosynthesis;

4 max [9]

M31. (a) (i) suitable reason for birth rate increase; examples, more people survive to reproductive age; better pre-natal care / health care of mother; better nutrition of mother;

1 max

(ii) suitable reason for death rate fall; examples, better nutrition; better sanitation; (widespread) introduction of health care; better post-natal care (mother or child); vaccination programmes;

1 max

(b) (i) birth rate decreasing; as the death rate constant but births minus deaths is falling;

2

(ii) reduces population growth until 1989/90 (as more (net) emigration); increases population growth from 1989/90 (as more (net) immigration);

2 [6]

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M32. (a) genetic variation/ variation in gene/allele(s) in populations for cyanide production; colder/below 0°C (January) areas, cyanogenic plants die in this cold/acyanogenic survive; non-cyanogenic allele/gene passed on more often/its frequency increases; warmer (January) areas cyanogenic plants at advantage, because of less herbivore selection pressure/feeding; so cyanogenic survive more often to pass on cyanogenic allele/gene.

4 max

(b) large (and equal) number of quadrats in each area; (reject several) random sampling method, described; (accept described ‘systematic’ method) percentage cover/point hits per quadrat/count plants; mean/average value for each area; statistics test to see if differences significant.

4 max [8]

M33. General Principles for marking the Essay:

Four skill areas will be marked: scientific content, breadth of knowledge, relevance and quality of language. The following descriptors will form a basis for marking.

Scientific Content (maximum 16 marks)

Category Mark Descriptor 16

Good

14 Most of the material reflects a comprehensive understanding of the principles involved and a knowledge of factual detail fully in keeping with a programme of A-level study. Some material, however, may be a little superficial. Material is accurate and free from fundamental errors but there may be minor errors which detract from the overall accuracy.

12 10

Average 8

Some of the content is of an appropriate depth, reflecting the depth of treatment expected from a programme of A-level study. Generally accurate with few, if any, fundamental errors. Shows a sound understanding of the key principles involved.

6 4

Poor 2 Material presented is largely superficial and fails to reflect the depth of treatment expected from a programme of A-level study. If greater depth of knowledge is demonstrated, then there are many fundamental errors.

0

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Breadth of Knowledge (maximum 3 marks)

Mark Descriptor 3 A balanced account making reference to most areas that might realistically

be covered on an A-level course of study. 2 A number of aspects covered but a lack of balance. Some topics essential

to an understanding at this level not covered. 1 Unbalanced account with all or almost all material based on a single

aspect. 0 Material entirely irrelevant or too limited in quantity to judge.

Relevance (maximum 3 marks)

Mark Descriptor 3 All material presented is clearly relevant to the title. Allowance should be

made for judicious use of introductory material. 2 Material generally selected in support of title but some of the main content

of the essay is of only marginal relevance. 1 Some attempt made to relate material to the title but considerable amounts

largely irrelevant. 0 Material entirely irrelevant or too limited in quantity to judge.

Quality of language (maximum 3 marks)

[25]

Mark Descriptor 3 Material is logically presented in clear, scientific English. Technical

terminology has been used effectively and accurately throughout. 2 Account is logical and generally presented in clear, scientific English.

Technical terminology has been used effectively and is usually accurate. 1 The essay is generally poorly constructed and often fails to use an

appropriate scientific style and terminology to express ideas. 0 Material entirely irrelevant or too limited in quantity to judge.

Guidelines for marking the essay

Introduction

The essay is intended for the assessment of AO4 (Synthesis of knowledge, understanding and skills) and Quality of Written Communication (Sections 6.4 and 6.5 in the specification). Examiners are looking for

• evidence of knowledge and understanding at a depth appropriate to A level

• selection of relevant knowledge and understanding from different areas of the specification

• coverage of the main concepts and principles that might reasonably be expected in relation to the essay title

• connection of concepts, principles and other information from different areas in response to the essay title

• construction of an account that forms a coherent response

• clear and logical expression, using accurate specialist vocabulary appropriate to A level

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Assessing Scientific Content

Maximum 16 marks. Descriptors are divided into 3 categories: Good (16, 14, 12), Average (10, 8, 6) and Poor (4, 2, 0). Only even scores can be awarded, i.e. not 15, 13, etc. Examiners need first to decide into which category an essay comes.

A good essay

• includes a level of detail that could be expected from a comprehensive knowledge and understanding of relevant parts of the specification

• maintains appropriate depth and accuracy throughout

• avoids fundamental errors

• covers a majority of the main areas that might be expected from the essay title (These areas will be indicated in the mark scheme). (Occasionally a candidate may tackle an essay in an original or unconventional way. Such essays may be biased in a particular way, but where a high level of understanding is shown a high mark may be justified.)

• demonstrates clearly the links between principles and concepts from different areas.

Note that it is not expected that an essay must be ‘perfect’ or exceptionally long in order to gain maximum marks, bearing in mind the limitations on time and the pressure arising from exam conditions.

An average essay

• should include material that might be expected of C/D/E grade candidates

• is likely to have less detail and be more patchy in the depth to which areas are covered, and to omit several relevant areas

• is likely to include some errors and misunderstandings, but should have few fundamental errors

• is likely to include mainly more superficial and less explicit connections

A poor essay

• is largely below the standard expected of a grade E candidate

• shows limited knowledge and understanding of the topic

• is likely to cover only a limited number of relevant areas and may be relatively short

• is likely to provide superficial treatment of connections

• includes several errors, including some major ones

Having decided on the basic category, examiners may award the median mark, or the ones above or below the median according to whether the candidate exceeds the requirements or does not quite meet them.

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Marking the essay

In marking scientific content, letters in the margin show each key area covered; these are used to assess the breadth of criteria. A single tick is used to indicate accurate coverage of each significant area, and a double tick to emphasise ‘good depth of content.’ Errors are indicated with a cross. A squiggly line in the margin is used to highlight irrelevance and ‘Q’ to highlight poor use of terminology, unclear grammar and inappropriate style.

Specific guidance for assessing Scientific Content and Breadth of Knowledge in Essays

The following provides guidance about topics which might be included in the essays. It is not an exclusive list; the assessment of scientific content does not place restrictions on topics that candidates might refer to, provided they are

• relevant;

• at an appropriate depth for A level and

• accurate.

It is not expected that candidates would refer to all, or even most, of the topics to gain a top mark; the list represents the variety of approaches commonly encountered in the assessment to the essays. In both essays, topics either from the option modules or beyond the scope of the specification should also given credit where appropriate.

Breadth of Knowledge

3 marks four topics - at least one from each set of examples 2 marks four from one set of topics three topics - at least one from each set 1 mark two topics

M34. (a) lactate/lactic acid/pyruvate; ATP; 2

(b) (i) energy demand is very high/high respiration rate; unable to supply enough oxygen to muscles/tissues/cells/ insufficient time for oxygen to reach muscles/tissues/cells / insufficient oxygen in muscles/tissues/cells;

2

(ii) break down with oxygen /oxidise lactate; convert to pyruvate / glucose / glycogen / CO

2 + water;

by aerobic respiration; 2 max

[6]

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M35. (a) lactate/lactic acid/pyruvate; ATP; 2

(b) (i) energy demand is very high/high respiration rate; unable to supply enough oxygen to muscles/tissues/cells/ insufficient time for oxygen to reach muscles/tissues/cells / insufficient oxygen in muscles/tissues/cells;

2

(ii) break down with oxygen /oxidise lactate; convert to pyruvate / glucose / glycogen / CO

2 + water;

by aerobic respiration; 2 max

[6]

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E1. (a) This question was the most effective discriminator on the entire paper. The best answers used all the information provided to describe how geographic isolation could cause a very high diversity of bird species. At the other extreme, speciation was ignored and a description of succession was given. Most answers did attempt to explain speciation but often did not make sufficient use of the information provided to gain high marks. Usually these accounts only gained the marks for geographic isolation and for describing differential reproductive success. Poor use of terminology was also clearly evident in these weaker responses. References to different selection pressures and changes in allele frequency were often only mentioned in better responses.

(b) Surprisingly, almost fifty percent of students failed to gain this mark. Common incorrect response referred to a climax community being formed, or mutations not occurring. Students gaining this mark often mentioned no ‘isolation’ or ‘a similar environment’.

E2. (a) About 50% of candidates gained both of the marks available for part (i), but of the rest

there was considerable evidence of confusion. Nearly all wrote out the equation p2 + 2pq +

q2 = 1, when finding 0.22

was all that was needed in this case. Many also did not know

whether the allele frequency of 0.2 was the value for q or for q2. Most candidates responded correctly to part (ii), but a number continued to provide irrelevant detail about the conditions required for the Hardy-Weinberg principle to be valid.

(b) Few candidates gained both the marks available for part (i), as they did not show the necessary understanding of the difference between chance and probability. The answer given by many to part (ii), stabilising selection, suggested that they had not read the stem of this part of the question carefully enough. Those candidates who missed marks in their explanations usually did so because they wrote generally about selection rather than explaining the effect of this allele on survival and reproductive success and the consequent decrease in its frequency.

E3. Candidates were not always selective in choosing the material to answer this question. They sometimes wrote at length about succession in part (a) (which concerned diversity), and then found they had nothing new to say in part (b), which did concern succession.

(a) Good candidates were able to state, clearly and unequivocally, that an increase in the index of diversity means that the number of species has increased, as has the number of individuals within each species. Weaker candidates sometimes wrote all they knew about succession here or sometimes just wrote about “an increase in the organisms“, failing to distinguish between new species and existing species.

(b) Most candidates were able to describe the role of pioneer species in colonising a harsh environment and the ways in which they might change this environment. Better candidates then went on to say that these changes allowed new species to become established with the creation of new habitats for still other species. However, too many could not resist following the development right through to the climax and describing the nature of this condition. Clearly, given the question, this was not necessary and could have wasted valuable time for some candidates.

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E4. Unit 6

Most candidates were able to cope with the genetics of colour blindness but the setting out of the genetic diagram often left much to be desired in terms of clarity of presentation. Those opting for the Punnett square approach typically omitted the parental genotypes and often failed to relate genotype and phenotype for the colour-blind male offspring. Those choosing to use line diagrams often omitted the gametes. A small minority did not understand proportions and could not distinguish between % and 1:4, or between 1:3 and 3:1. Some multiplied their correctly derived answer by lA to allow (again) for the child being male.

Unit 7

It was pleasing to see that many candidates could produce a genetic diagram that enabled them to score of at least 2 marks, with better candidates scoring maximum credit. The parental genotypes and the probability value were usually correct, although several omitted the parental genotypes when using a Punnett square. Common errors included omission of the gametes where direct lines between parental and offspring genotypes were used, or a failure to identify the male with red-green colour blindness. Giving a ratio of 1:4 or 3:1 and stating the male

parent’s genotype as XrY were less frequent but still quite common errors. Several candidates multiplied their correct probability of % by lA because they failed to realise they had already taken account of the child being male.

E5. (a) Most candidates were able to score quite well here, but too many just produced a standard response to the question “describe the light-independent reactions of photosynthesis”. However, the question required candidates to explain how these reactions allow the continued synthesis of hexose, which demanded that they make use of the information in the diagram.

(b) There were many excellent descriptions of the role of the electron transport chains in the light-dependent reactions in producing both ATP and reduced NADP. However, some candidates did not give any account of the production of reduced NADP, and some confused it with NAD. Others unfortunately described the synthesis of ATP in the electron transport chains involved in oxidative phosphorylation and, consequently, were able to score very few marks.

(c) Although a good number of candidates realised that some of the hexose produced in photosynthesis is used in respiration, only a few could explain that this resulted in mass loss due to the loss of carbon dioxide. Many could not separate mass and energy and suggested that because energy is released in respiration, this accounted for the mass loss. A few candidates realised that, over a twelve-month period, parts of the plant may be lost to decomposers and parts may be eaten by animals, both of which would reduce the increase in dry mass over that period.

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E6. Unit 8

(a) Most candidates were able to explain that an increase in water temperature would influence a relevant feature such as oxygen solubility or respiration. Answers based on the effect of temperature on the rate of enzyme activity or on metabolism were, however, a little too general, failing to relate to the specific investigation described in the question. There were a few references to ensuring “a fair test”, an entirely inappropriate response at this level.

(b) The responses to this section formed a sharp contrast to the high marks frequently awarded for statistical analysis in coursework. Answers to part (i) were often centre-dependent, some candidates being able to produce a sound null hypothesis; others clearly had little idea. These candidates frequently lacked understanding of the purpose of the investigation or of the concept of a null hypothesis. The weakest responses usually involved equating the expression with an inappropriate statistical formula. In part (ii), many candidates were aware that statistical tests are related to chance, but fewer were able to explain that such tests give a measure of the probability that chance might account for the results obtained.

(c) Most candidates correctly identified A as the more likely explanation and were able to justify their choice.

(d) Better candidates were able to produce in a logical account in which they successfully linked a lower oxygen concentration to anaerobic respiration and the production of lactic acid. Others revealed a disturbing lack of understanding of respiratory biochemistry, suggesting that the evolution of carbon dioxide was entirely independent of the consumption of oxygen. They inevitably based their answers on an argument that, despite reduced oxygen, fish must continue to respire aerobically, so there would be an increase in carbon dioxide. There were occasional references to supposed chemical effects of zinc.

(e) The best candidates used common sense in part (i) and, realising that the only elements that were concentrated were copper and cadmium, calculated appropriate ratios for these ions. Credit was also given to those who supported their conclusions by calculating the inverse. A significant number, however, merely subtracted the relevant values from each other, an approach which inevitably led to an incorrect answer. The examiners were instructed to be generous in marking the calculations and undertook much work in interpreting confusing presentation. Centres would do well to advise candidates that it is their responsibility to present material sufficiently clearly that the logic of the response can be followed. In part (ii), most recognised that lead ions would be egested or excreted, although there was some incorrect usage of these terms. Most candidates were aware, in part (iii), that woodlice would concentrate copper. The principle of bioaccumulation was often correctly described but not always related to eating a large number of leaves. Weaker candidates frequently referred to additional sources of copper ions or to the intriguing possibility of copper ions multiplying within the body of the woodlouse.

(f) Mutation figured widely in the responses to part (i), although there were occasional incorrect references to natural selection or to the presence of arsenic causing the allele to first arise. Although there were a number of rather vague references to growth and formation of new cells, the majority of candidates were able to identify two specific effects of phosphates in part (ii). Answers to part (iii) were frequently marred by a failure to answer the question and explain why arsenic-tolerant plants were unable to compete in the conditions described. Candidates referred to both arsenic-tolerant and non-tolerant plants as “they” and it was often far from clear as to which they were referring. However, it was encouraging to note that, although this question was targeted specifically at Grade A candidates, many others were able to suggest that arsenic-tolerant plants would not grow as well because they were unable to take up sufficient phosphates.

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Unit 9

(a) Candidates offered a range of explanations, suggesting that the air mixed the water, that it affected the zinc, and that it was needed to make the test fair. Few candidates earned a mark; those that did suggested that oxygen would no longer be a limiting factor. Links were rarely made with the effect it would have on the saturation of the haemoglobin.

(b) (i) A large number of individuals know from their coursework that the term ‘null hypothesis’ implies ‘no difference’, but they did not always recognise where this lack of difference might lie. Weaker candidates made comments about chance. The commonest error was to devise a hypothesis relating to gas exchange and respiration.

(ii) Many commented on the need to look for effects that are due to chance. Some quoted significance levels, but failed to mention probability. Many referred to establishing levels of accuracy, and a few made statements about the null hypothesis. It was disappointing to note that large numbers of candidates are able to suggest null hypotheses in their coursework but are unable to apply these statistical skills to material presented in an unfamiliar context.

(c) Most candidates recognised the answer as A, and were able to use the graph to explain their choice. Those that could not were vague in their answers.

(d) Unless candidates recognised that there was less oxygen available to the cells they were inclined to answer irrelevantly. The best recognised the anaerobic respiration that would ensue, and therefore lactic acid would be produced. Some wrote of haemoglobin as a buffer, but failed to recognise that it would be the extra hydrogen ions which affected the pH not those absorbed through the buffer. Weaker candidates were confused over the numbering of the pH scale. They thought that zinc affected the pH of the water, or that zinc caused haemoglobin to pick up fewer hydrogen ions from the water.

(e) (i) The calculations were absent in some cases, and very varied where present. Simple ratios were the best idea, but some even calculated standard deviations. Subtractions were also fine. Many candidates had no idea what to calculate. The commonest response was to find the mean concentration of cadmium and copper in shrews, without any reference to the levels in the source of food. Many gave calculations without saying what they were, leaving the examiner to guess. The weakest candidates mis-read the data as numbers of shrews or numbers of ions. Despite poor performances on the supporting mathematics, many candidates could comment on the relative concentrations.

(ii) The fate of the ions was mixed. There appears to be widespread confusion over egestion and excretion and the fact that ions have to be absorbed before they can be used appears to have escaped some. Weaker candidates were of the opinion that the copper ions could be broken down.

(iii) Candidates had little understanding of the ways in which ions accumulate through diet.

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(f) (i) Most candidates correctly identified mutations as the cause of the allele arising, but some offered a choice to the examiner regarding natural selection.

(ii) Likewise, most candidates were able to name two functions of phosphates. A few were confused with protein synthesis. Some answered too vaguely with ‘membranes’, iii). This part of the question presented difficulties to many and only the better candidates directed their responses appropriately. There were many vague references to “fogs” and inappropriate set-piece answers on inheritance.

E7. (a) A majority of candidates were able to score all three marks available for this section. Nearly all candidates derived correct parental genotypes, although a few made a fundamental error in giving what was, effectively, a haploid genotype (AB, rather than AaBb). Some then lost their way because they wrote down four gametes for each parent - two of each genotype. They then produced a 4 x 4 Punnett square. Because previous experience of this had always produced a 9:3:3:1 ratio, they assumed it would here also.

(b) Again, a majority produced the correct answer of a frequency of 0.6 for the a allele. Some candidates, however, correctly identified the combined frequencies of the rose and single-

combed chickens as q2, with a frequency of 0.36. They then obtained q by taking the square root of this to obtain 0.6. Perhaps because they have often been required to find the frequency of the heterozygote, they did this next and offered 0.48 as their answer. This is another instance where not reading the question carefully may have cost marks.

E8. (i) Most candidates knew the meaning of the term "population". However, imprecise language, as ever, cost marks. A population is all the organisms of one species in a particular area (at a particular time). It is not all the organisms of a species, or a group of organisms of one species in a particular area.

(ii) Most candidates explained that deforestation destroys habitats and this reduces the prey numbers/biomass. However, few used the graph to say that only if the prey biomass fell below 600 tonnes per 100 km2 would the tiger population decrease.

E11. (a) Nearly all candidates knew that habitats would be lost, and a good number also knew that this would reduce the number of species in the area.

(b) Better candidates realised that both potassium nitrate and ammonium nitrate would release nitrate ions into the soil immediately, and that the ammonium ions in ammonium nitrate could be nitrified by nitrifying bacteria to provide a secondary release of nitrates. Many candidates, however, just did not apply their knowledge to the problem and merely recited chunks of the nitrogen cycle, with some confusion between nitrifying and nitrogen-fixing bacteria.

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E12. Many candidates from across the ability range were able to complete the genetic diagram and gain full marks. Some did not realise the dihybrid nature of the cross and some did not always make clear the relationship between offspring genotypes and phenotypes.

E13. (a) Nearly all candidates realised that deforestation would lead to loss of habitats, and many also realised that, as a result, some species would not survive in the area. Some missed the second point writing that ‘organisms’ would move out of the area. This need not lead to a reduction in diversity.

(b) Parts (i) and (ii) showed that most candidates were aware of the roles of photosynthesis, respiration and combustion in the balance of carbon dioxide in an area and how these might (or might not) change following deforestation.

E14. (a) Most candidates were able to explain the meanings of the two terms correctly, but again, a lack of precision cost marks for some candidates. It is insufficient to describe a population as ‘a group of organisms of the same species’ or ‘all the organisms of a species’. It is, similarly, insufficient to describe a community as ‘ a group of populations’.

(b) (i) Most candidates were able to quote suitable conditions necessary to ensure the validity of the mark-release-recapture technique. They were also usually able to calculate the size of the woodlice population.

(ii) Most candidates were able to quote suitable conditions necessary to ensure the validity of the mark-release-recapture technique. They were also usually able to calculate the size of the woodlice population.

(c) Most candidates knew that the quadrats must be placed randomly and many were able to describe a method of achieving this. They usually realised that the number of dandelion plants per quadrat must be counted (although some suggested estimating percentage cover, which is not suitable in this instance) and the count repeated. However, rather fewer went on to say that one could then calculate the mean number per quadrat and, from this, estimate the number in the field by multiplying by the ratio of area of field to area of quadrat.

(d) (i) A majority of candidates was able to interpret the unfamiliar diagram to establish both the niches of the two species and the areas from which they were excluded. Some established the ‘exclusion areas’ correctly but not the basic niches.

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(ii) Some candidates confused sympatric and allopatric speciation, but the majority answered along the right lines. Most were able to establish the principle of reproductive isolation and could usually suggest a suitable mechanism that would bring this about. However, candidates frequently confused species and populations in their answers, which led to confusion about when the processes they were describing had occurred. For instance, many wrote about ‘gene pools of the two species becoming more and more different until they could no longer interbreed’ when what they should have been describing was gene pools of the populations. However, a good number established the principle that the two would be distinct species when they could no longer interbreed to produce fertile (rather than viable) offspring.

E15. (a) Too many candidates saw two empty boxes in the flowchart in (i) and either wrote the names of both substances in the boxes or the number of carbon atoms in each substance. This clearly is the result of not reading the question carefully. Those who did answer the question set, usually scored both marks. In part (ii) good candidates realised that all ATP is produced in mitochondria, except that produced in glycolysis. They therefore arrived at the correct answer of 36 ATP by deducting 2 from the net total yield of 38 ATP per molecule of glucose, or by deducting 4 from the total production of 40 ATP. Others did arrive at the correct answer by working out where each molecule of ATP was produced, but many attempting this method did so in a disorganised way and so made errors in calculation. In (iii) most candidates knew that, in the absence of oxygen, some of the reactions of respiration could not take place, but many were unable to describe the extent of anaerobic respiration. Well prepared candidates were able to state clearly that only glycolysis would take place and, therefore, the ATP production of the Krebs cycle and electron transport chain would be lost. They also often

(b) Despite being given specific information in part (i) concerning the features of the heterocysts (thick walls and the absence of chlorophyll), and the requirements of nitrogen fixation (anaerobic conditions) candidates too often invented other features and reasons other than maintaining anaerobic conditions for those features. Disappointingly few candidates confined themselves to answers based on excluding oxygen and not producing oxygen, which would inhibit the process of nitrogen fixation. There were some excellent answers to part (ii) from candidates who appreciated that nitrogen-containing compounds in the rice plants would be the starting point for the reactions of the nitrogen cycle, and duly described the roles of decomposition and nitrification accurately and logically. Some realised that the decomposers would produce carbon dioxide as a result of their respiration and that this could be used in photosynthesis by the leaves of the rice plants. However, too many just assumed that the ammonia produced by the heterocysts would be released into the soil, apparently unused by the fern and, in their answers, took this as the starting point for the nitrogen cycle. This clearly shows less appreciation of the situation as described.

E16. (a) Nearly all candidates knew that members of the same species can reproduce to produce fertile offspring. Fewer, made the additional point that they shared similar features.

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(b) Responses to this question were disappointing and were most likely due to candidates not really looking carefully at the evidence and realising that the means and ranges of beak depth of the two species on island 3 had shifted in opposite directions. A majority of candidates thought that the changes in distribution of beak depths of the two species on island 3 were an example of disruptive selection. It is likely that these candidates focused on the third graph, saw two distributions and assumed that there had been selection in favour of the two extremes of one distribution. Some even went as far as to say that sympatric speciation had occurred.

E17. (a) Most candidates clearly appreciated that the samples would best be obtained by using random numbers to determine co-ordinates, although they were not always specific as to how these numbers would be generated. There were, however, frequent references to “throwing” a quadrat, a technique which does not give rise to a genuinely random distribution.

(b) In part (i), candidates revealed considerable difficulty in recognising the trend of a decrease followed by stabilisation from a depth of around 200 cm. The most frequent response was to ignore the change in gradient and refer simply to the population falling. Where a genuine attempt was made to offer an explanation for the difference in numbers in part (ii), answers were usually correctly related to the decreasing concentration of oxygen available for respiration.

(c) There was obvious confusion in the minds of some between the concepts of ecological succession and natural selection. While some of those who made this distinction produced excellent answers, others lost their way in rambling anthropomorphic accounts of bacteria “not finding conditions to their liking” or being “happier” with conditions at other depths. There were also many general references to “bacteria”. These lacked the necessary precision to gain significant credit.

(d) Part (i) was generally well answered and there were many accounts based on correct references to the surface layers being the only ones where numbers of aerobic bacteria increased. Part (ii) also produced some sound responses although candidates were inclined to embellish their answers with irrelevant detail relating to the anticipated change with time. Once again, a failure to gain marks most commonly stemmed from imprecise use of the word “bacteria”. In both parts (c) and (d) there was a need to refer them as being either aerobic or anaerobic.

(e) There was encouraging evidence of a good understanding of standard error and many candidates were able to show some appreciation of this demanding concept. However, answers were seldom targeted at explaining what the error bars revealed about the difference in activity at the times given. In spite of the wording of the last sentence of the question, the terms probability and chance were seldom incorporated into the explanation.

(f) The most frequent approach to this part of the question was to produce a mass of figures supporting a complex but inappropriate calculation. With these data the best approach was to construct a graph and draw a line of best fit. The likely number of bacteria could then have been established by reading off the appropriate value from the curve.

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E18. Fertilisers are well known and understood by most candidates. However, in part (a), candidates often failed to gain marks through vague answers, such as ‘cheaper’ without qualification, or by being too dogmatic; for example, “manure does not cause leaching or eutrophication”. Too many were concerned about the smell of manure. In (b) (i), there were still many candidates who did not understand the concept of a control plot. Many suggested using both fertilisers, or one followed by the other. In (b) (ii), many candidates realised that the inorganic fertiliser may have contained a nutrient which mangolds required.

E19. Although this question produced a wide range of marks, few candidates gained maximum marks, often due to an inadequate explanation in part (b) (iii).

(a) Few candidates obtained all four marks. Most candidates gained two marks, usually for identifying where NADP is reduced and where ATP is produced. A common error was to indicate NAD is reduced in the light-dependent reaction of photosynthesis.

(b) (i) Most candidates incorrectly suggested glucose as a substrate for this investigation. However, there was a significant number of correct answers usually suggesting pyruvate or acetylcoenzyme A.

(ii) Many candidates referred to the phosphorylation of ADP to produce ATP but the fate of oxygen was less well known. A common misconception was to suggest that oxygen is used in the production of carbon dioxide.

(iii) Although some candidates gave the correct order of the electron carriers, many candidates got the order the wrong way round. Very few candidates could provide an adequate explanation although there were some excellent exceptions to this. It was not uncommon for candidates to simply describe the electron transport chain in mitochondria.

E20. (a) In (i), almost everybody sketched the pyramid correctly. Some did not draw the pyramid in the conventional way; in effect they drew it upside down, but still showed the correct relationships between the trophic levels. A few candidates failed to the gain the mark because they did not label their diagram. Likewise, in (ii), nearly everyone knew that energy is lost between trophic levels and could suggest at least one way in which it is lost.

(b) In this question, it was a relatively common failing for candidates to be unable to write about the relevant part of the nitrogen cycle without also trying to include other aspects. In (i), some candidates failed to distinguish between the roles of decomposers and nitrifying bacteria in their answers, phrasing their responses along the lines of “the decomposers and nitrifying bacteria convert the organic substances into ammonia and then to nitrates”. It was also disappointing to read the number of answers that included references to lightning and the Haber process. However, candidates who understood the nitrogen cycle well usually had little problem with this question. There was a general understanding in (ii) that nitrogen-fixing bacteria convert nitrogen gas into a form that is more readily available; however, there was also a widespread misconception that they convert the gas directly into nitrate ions.

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(c) In (i), a disappointing number of candidates did not read the question carefully and described changes in the populations of both types of bacteria and both types of protoctistans, usually without really explaining the reasons for any of the changes. Good candidates recognised the predator-prey relationship between the dispersed bacteria and the free-swimming protoctistans in the way the numbers increased and then declined slightly out of phase with each other. In (ii), candidates who understood the process of succession were generally able to recognise the changes in the environment in the treatment tank that resulted in changes in the community inhabiting that environment. A common failing was not to make clear that it is the activities of the organisms that inhabit an area that change the environment and so make it suitable for colonisation by other species.

E21. This question produced the full range of marks, with some excellent answers and some with little knowledge of the nitrogen cycle and its importance to farming.

(a) Many candidates scored full marks, but there were also many who seemed to be naming any bacterium which was part of the nitrogen cycle.

(b) Only a minority of candidates scored full marks. There were many incomplete answers from candidates who just named a farming practice, and then gave no detail of the nitrogen compounds involved or how the farming practice would add or remove the nitrogen compounds. Many candidates did not seem aware of what constituted a farming practice, with many references to adding various types of bacteria or water-logging fields, or describing totally unrelated practices such as spraying pesticides. ‘Growing legumes’ and ‘hedge removal’ with associated explanations were the most common correct answers to (i) and (ii) respectively. Many candidates thought that the growing of any crop, and then ploughing it, would increase the nitrogen content.

E22. The question was generally well answered with the majority of candidates scoring more than half marks. However, only a minority of candidates scored full marks.

(a) The majority of candidates scored two marks for describing the importance of photosynthesis and respiration. Only a few candidates developed their answer fully to explain the role of enzymes in the breakdown of organic matter by bacteria, or explained how algae use nitrates and ammonia to form organic nitrogen compounds. Most just stated that they are needed for ‘growth’. The weaker candidates tended to describe the information given in the diagram rather than explain it.

(b) The majority of candidates linked the increased temperature to higher enzyme activity, but many of these failed to develop their answer further to discuss the effect of this on named metabolic processes. Many candidates just referred to enzymes working more efficiently or stated that the enzymes would be at their optimum. Many also stated that the shallow pond was to keep the bacteria and algae close together.

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E23. (a) Most correctly pointed out that sex-linkage was indicated in the pedigree diagram by the fact that the condition was seen only in males. Many made the false assumption that being expressed by the minority demonstrated that the allele was recessive; better candidates realised that if the trait was not expressed in the parents but did appear in the offspring then its allele must have been recessive. Some were non-selective in their use of evidence and forfeited the mark. Many insisted on including underlying theoretical points which were not necessary as they did not constitute the evidence (i.e. observations) required by the question.

(b) Most candidates identified the possible genotypes of all four individuals correctly, including the two possibilities for person 7. Unfortunately, many were rather careless with the final genotype, assuming it to be the same as for number 7 (i.e. disregarding the extra information given in the diagram).

(c) Many calculated the probability correctly as 0.25, often using the extra space available on the page to draw genetic diagrams to verify their ideas. Some spoiled their answers by giving extra, incorrect alternatives – thus, ‘25% or 1:4’ scored zero. Others did not appreciate the difference between ‘1:3’ (correct) and ‘3:1’ (incorrect).

E24. Poor expression and a tendency to describe rather than explain caused the majority of candidates to score poorly on this question.

(a) The majority of candidates described the pattern shown by the graph. Of those who did explain, the answers were often incomplete. ‘Intraspecific’ competition was rarely mentioned, and candidates often referred vaguely to ‘resources’, rather than a named resource.

(b) Candidates also tended to state rather than explain, and so could not be awarded the mark. In part (ii), poor expression and incomplete answers were common, especially when expressing the chance of survival of the chicks or the reason why chicks were unlikely to develop into adults and pass on their genes. A significant number misread the question and explained why 14 rather than 3 chicks would be an advantage.

E25. Generally this question was well answered with most candidates obtaining at least five marks. However, part (d)(i) proved difficult for a significant number of candidates.

(a) This caused few problems with the vast majority of candidates correctly explaining that a dominant allele is always expressed in the phenotype or codes for a functional protein.

(b) Most candidates correctly named the relationship between the two alleles as codominance. A common incorrect response was epistasis.

(c) The majority of candidates had little difficulty completing the genetic diagram to obtain all four marks. However, a number of candidates failed to gain a mark for the correct ratio of offspring phenotypes. Candidates failing to gain any marks often attempted a monohybrid cross.

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(d) (i) Only better candidates gained both marks. Common incorrect responses referred to mutations or to sperm being XX or XY.

(ii) Although many candidates did refer to obtaining milk or meat, not all candidates linked this to the gender of the cattle. It was disappointing to find a significant number of A level biologists referring to ‘milk from bulls’.

E26. (a) This was poorly answered as few candidates referred to the loss of high protein parts of the plants in dry conditions or the reduced uptake of nitrates. Most stated that, as the process of photosynthesis required water, there would be less energy or less carbohydrate for protein synthesis.

(b) Only a minority of candidates correctly explained the higher protein content in the food by selective eating of the high protein species/parts. Many stated that wildebeest had another source of protein, such as animals.

(c) There was a very wide variety of acceptable answers, although most candidates described the effect of a lack of actin or myosin. A significant number of candidates gave non-protein examples such as glucose, starch and acetylcoenzyme A or did not name a protein.

E27. Failure to use the correct ecological terms was a major problem in this question.

(a) Succession was only explained well by a minority of candidates. The term ‘community’ was seldom used – instead candidates referred to one species replacing another, or just stated that the ‘plants’ or ‘organisms’ would change. Many candidates described succession purely in terms of an increase in the number of species. Specific changes in environmental factors or the role of species in changing these factors were seldom mentioned. Climax community was described more accurately with most candidates stating that it was the end point of succession.

(b) Most candidates correctly explained the changes in the number of other species in terms of interspecific competition for light or nutrients. There were many vague references to ‘resources’ or ‘space’ which could not be given credit.

(c) Only a minority of candidates answered in terms of biomass production, i.e., the balance between photosynthesis and respiration. Most did not use the information given in the table about the appearance of the heather, and just discussed the changes in terms of competition.

E28. (a) This was poorly answered as few candidates referred to the loss of high protein parts of the plants in dry conditions or the reduced uptake of nitrates. Most stated that, as the process of photosynthesis required water, there would be less energy or less carbohydrate for protein synthesis.

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(b) Only a minority of candidates correctly explained the higher protein content in the food by selective eating of the high protein species/parts. Many stated that wildebeest had another source of protein, such as animals.

(c) There was a very wide variety of acceptable answers, although most candidates described the effect of a lack of actin or myosin. A significant number of candidates gave non-protein examples such as glucose, starch and acetylcoenzyme A or did not name a protein.

E29. (a) This was not well answered by many candidates. Most knew that legumes fixed nitrogen but references to roots, nodules or bacteria were often missing or incorrect. Very few discussed the advantage in a crop rotation, and just discussed the need for less fertiliser.

(b) Most gained one mark for stating that applying more than 600 kg of fertiliser did not increase yield, but many failed to use the data or state that lower rates of application led to an increase in yield.

(c) This was either well answered or very poorly answered. If candidates made the link between fertiliser and water potential, they usually scored both marks but many candidates were unable to make this link.

E30. (a) This discriminated well, better candidates gained both marks but weaker ones simply stated it was bioaccumulation, showing little understanding of why it occurs.

(b) This was well answered and most candidates were able to suggest a reason for leaving a strip of bare ground between the hedgerow and the conservation headland.

(c) This question was generally poorly answered with vague responses and poor terminology. Only the best candidates discussed ideas like habitats for pest predators. Weaker candidates refered to ‘homes for insects’, or stated increased diversity with no explanation.

(d) This was well answered by many, with candidates gaining three marks for a discussion of the nitrogen cycle. Few candidates discussed the carbon cycle, and those that did were rather vague.

E31. Good candidates scored well on this question but only the very able gained maximum marks.

(a) (i) Few candidates scored here, with a lack of contraception, increased IVF availability and soldiers returning from war being the most favoured incorrect responses. If food availability was mentioned, then it was not clearly linked with the female.

(ii) This was well answered, usually referring to better health care or food availability. There were some candidates who did not read the question and stated that the decrease in death rate was actually due to an outbreak of a virulent disease.

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(b) (i) This had an easy mark for pointing out that the birth rate decreased, which many candidates gained, but only the better candidates could use the information provided in the graph with any clarity. Many candidates did not score the second mark not through a lack of knowledge, but rather an inability to follow the rubric of the question and clearly use the (Births-Deaths) line to explain their answer, also pointing out that the death rate was constant.

(ii) This proved to be a discriminating question with candidates who used the dates (before, and after, 1989/90) correctly scoring both marks. Some candidates attempted, incorrectly, to incorporate the birth and death rate data into the emigration and immigration data to give an overall effect on the population.

E32. Many good answers were seen to both parts of this question. The topics covered were obviously familiar to many candidates. Where marks were not gained, it was usually because of omissions rather than errors. The full range of marks was seen and the question discriminated well.

(a) Almost all of the candidates obtained a mark for noting that cyanogenic plants might die in areas with very low mean January temperatures. Many went on to obtain a second mark for identifying the positive advantage that cyanogenic plants have in warmer areas, because they deter herbivores. Only the better candidates wrote about the impact of these different selection pressures on allele frequencies in different clover populations. It was encouraging to note that ‘rote answers’, unrelated to this example were absent. Some weaker candidates did fail to score marks because they wrote in general terms about factors affecting natural selection and evolution but with no reference to the specific factors given in this example.

(b) The vast majority of candidates were familiar with the use of quadrats. Many were also able to describe a suitable method for placing these at random locations in the study areas. Some candidates suggested the use of transects and this suggestion was rejected; unless they suggested the use of very large numbers of transects along randomly chosen lines. Many candidates scored one mark for suggesting the use of large numbers of quadrats. A surprising number failed to get this mark, either because they made no reference to sample size, or because they wrote vaguely about ‘several’ quadrats being used. The majority of candidates obtained a mark for what a quadrat might be used to measure. A large number of candidates made reference to the use of statistics but often that was all they said. The examiners were looking for the use of a statistical test to determine whether or not there was a significant difference in the number of clover plants in the two areas.

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E33. Both essays appear to have been attempted by about equal numbers of candidates. The full range of marks was seen for both essays. To obtain high marks, candidates had to use scientific terminology and use it correctly.

Some candidates produced excellent essays. These essays contained good, relevant detail about the stages in photosynthesis where carbon dioxide is fixed and in respiration where carbon dioxide is produced. They also gave good accounts of the transfer of carbon compounds through food chains, using the appropriate terminology. This was usually linked to detailed descriptions of the carbon cycle. Many good essays included detailed accounts of some of the following; digestion and the role of enzymes, how carbon-containing substances move through cell membranes, the transfer of DNA during reproduction and the exchange of carbon dioxide across exchange surfaces. Many candidates gave detailed accounts of only one or two topics and then resorted to generalised accounts of others. The weakest candidates usually gave very generalised accounts of the carbon cycle, at a level comparable to GCSE. Quite a large number of candidates included irrelevant material about global warming and eutrophication.

E34. (a) Most candidates scored full marks, identifying lactate and ATP as the products. Carbon dioxide was a common error.

(b) Most of the marks missed in this question were in part (i) with candidates failing to explain why muscle tissue becomes deprived of oxygen or to relate the high energy demand to respiratory rate. A frequent approach was an explanation of anaerobic respiration as a rapid process, which was not credited. Part (ii) was well answered.

E35. (a) Most candidates scored full marks, identifying lactate and ATP as the products. Carbon dioxide was a common error.

(b) Most of the marks missed in this question were in part (i) with candidates failing to explain why muscle tissue becomes deprived of oxygen or to relate the high energy demand to respiratory rate. A frequent approach was an explanation of anaerobic respiration as a rapid process, which was not credited. Part (ii) was well answered.

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(5) (1) (total 6 marks) - the student room

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