STRETCH QUESTIONS - UNIT 4

175 minutes

140 marks

Page 1 of 51

Q1.          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.

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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Q2.          (a)     Explain what is meant by the ecological term community

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(b)     Scientists investigated the distribution of three species of fish in a lake. They recorded the range of depths where each species was found. The table shows their results.

 

(i)      Use information from the table to give the range of depths at which all three species of fish may be found living together.

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

Species of fish Range of depths /m

White bass      0 to 8.4

Walleye 6.8 to 10.0

Sauger 7.2 to 14.6

(ii)     Suggest and explain one advantage to the fish of occupying different depths in the lake.

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(c)     The graph shows the relationship between the depth and the temperature of the water in the lake.

A student concluded that the temperature of the water in the lake determined the depth at which the species of fish were found. Use the table and the graph to evaluate this conclusion.

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(Total 7 marks)

   

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Q3. (a)     Rather than use chemical pesticides or biological agents farmers often use an integrated system of chemical pesticides and biological agents to control agricultural pests. Explain the advantages of using an integrated system to control agricultural pests.

(6)

(b)     Nitrate from fertiliser applied to crops may enter ponds and lakes. Explain how nitrate may cause the death of fish in fresh water.

(5)

(c)     Explain how the intensive rearing of domestic livestock increases net productivity. (4)

(Total 15 marks)

Q4. Scientists investigated the effect of a pesticide called malathion on the survival of tadpoles of species of toads found in the USA. The scientists determined the LC50 for the tadpoles of each species over a 16-day period in an aquarium. The LC50 is the concentration of malathion that killed 50 percent of a population of tadpoles. The scientists also investigated whether the presence of a predator of tadpoles changed the effect of malathion.

(a)     Suggest two advantages of using the LC50 to determine the effect of a pesticide.

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The scientists looked at previous studies on the effects of various pesticides on tadpoles.

They found that most of these studies:

•        were carried out on tadpoles of the African clawed toad

•        measured the LC50 of each pesticide over 1 to 4 days in the absence of any biotic factor.

(b)     The scientists concluded that these previous studies were of limited use when trying to assess the effects of malathion on the tadpoles of toads found in the USA. Suggest why the scientists reached this conclusion.

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(c)     Malathion affects the nervous system of tadpoles. The scientists investigated whether the stress caused by the presence of a predator changed the effect of malathion on the tadpoles. The scientists used newts which are predators of tadpoles. They carried out two experiments, as shown in the diagram.

 

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(i)      Explain why the scientists carried out experiment 1.

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(ii)     Explain why the scientists used a transparent barrier in experiment 2.

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(d)     Adult toads spend most of their time on land but lay their eggs in water.These eggs hatch into tadpoles, which live in water and develop into adults. The tadpoles are much smaller than adult toads. Use this information to explain why the tadpoles are affected more rapidly by pesticides in water than adult toads.

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(e)     When malathion is used as a pesticide, it is often sprayed onto aquatic habitats at

concentrations of 0.1 to 1.6 mg dm–3. The scientists tested the effect of malathion at concentrations of 0.001 to 10 mg dm–3. Suggest why.

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(f)     As a result of this investigation, the scientists concluded that more studies on pesticides should be carried out in natural habitats rather than under laboratory conditions. Suggest two advantages of carrying out such investigations in natural habitats.

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(Total 15 marks)

Q5. Yeast is a single-celled organism. A student investigated respiration in a population of yeast growing in a sealed container. His results are shown in the graph.

 

(a)     Calculate the rate of oxygen uptake in arbitrary units per hour between 2 and 4 hours.

 

 

 

Answer ...................................... arbitrary units per hour

(1)

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(b)     (i)      Use the information provided to explain the changes in oxygen uptake during this investigation.

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(ii)     Use the information provided to explain the changes in production of ethanol during this investigation.

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(c)     Sodium azide is a substance that inhibits the electron transport chain in respiration.The student repeated the investigation but added sodium azide after 4 hours. Suggest and explain how the addition of sodium azide would affect oxygen uptake and the production of ethanol.

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(Total 9 marks)

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Q6.          (a)     The biochemical pathway of aerobic respiration involves a number of different steps.

Name one step in which carbon dioxide is produced.

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In an investigation, scientists transferred slices of apple from air to anaerobic conditions in pure nitrogen gas. They measured the rate of carbon dioxide production.

(b)     The scientists kept the temperature constant throughout the investigation. Explain how a decrease in temperature would affect the rate of carbon dioxide production.

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(c)     When the apple slices were transferred to nitrogen, the following biochemical pathway took place.

Use this pathway to explain the part played by reduced NAD when the apple slices were transferred to nitrogen.

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(d)     The rate of carbon dioxide production was higher when the apple slices were in nitrogen than when they were in the air. Explain why.

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(Total 8 marks)

   

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Q7. Scientists investigated the effect of iron deficiency on the production of triose phosphate in sugar beet plants. They grew the plants under the same conditions with their roots in a liquid growth medium containing all the necessary nutrients. Ten days before the experiments, they transferred half the plants to a liquid growth medium containing no iron. The scientists measured the concentration of triose phosphate produced in these plants and in the control plants:

•        at the end of 6 hours in the dark

•        then for 16 hours in the light.

Their results are shown in the graph.

(a)     (i)      The experiments were carried out at a high carbon dioxide concentration. Explain why.

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(ii)     Explain why it was important to grow the plants under the same conditions up to ten days before the experiment.

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(iii)    The plants were left in the dark for 6 hours before the experiment. Explain why.

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(b)     Iron deficiency reduces electron transport. Use this information and your knowledge of photosynthesis to explain the decrease in production of triose phosphate in the iron-deficient plants.

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(c)     Iron deficiency results in a decrease in the uptake of carbon dioxide. Explain why.

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(Total 9 marks)

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Q8.          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.

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(Total 7 marks)

   

Q9. (a)    Explain what is meant by the term phenotype.

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(b)     Tay-Sachs disease is a human inherited disorder. Sufferers of this disease often die during childhood. The allele for Tay-Sachs disease t, is recessive to allele T, present in unaffected individuals. The diagram shows the inheritance of Tay-Sachs in one family.

(i)      Explain one piece of evidence from the diagram which proves that the allele for Tay-Sachs disease is recessive.

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(ii)     Explain one piece of evidence from the diagram which proves that the allele for Tay-Sachs disease is not on the X chromosome.

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(c)     (i)      In a human population, one in every 1000 children born had Tay-Sachs disease. Use the Hardy-Weinberg equation to calculate the percentage of this population you would expect to be heterozygous for this gene. Show your working.

 

 

 

 

 

Answer = ...................................... % (3)

(ii)     The actual percentage of heterozygotes is likely to be lower in future generations than the answer to part (c)(i). Explain why.

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(Total 10 marks)

Q10.          (a)     Explain what is meant by stabilising selection and describe the circumstances under which it takes place.

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(b)     Some European clover plants can produce cyanide. Those plants that can produce cyanide are called cyanogenic; those that cannot produce cyanide are called acyanogenic. Cyanide is toxic to the cells of animals and plants.

When the leaves of cyanogenic plants are damaged by slugs, or exposed to low temperatures, membranes within the cells are broken. This causes the release of the enzymes that control the reactions which produce cyanide.

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The proportions of cyanogenic and acyanogenic plants in clover populations were determined in different parts of Europe. These are shown in the diagram below, together with the mean minimum winter temperatures. Slugs are not usually active at temperatures below 0 °C.

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Explain the proportions of cyanogenic and acyanogenic plants in clover populations growing in the area where the mean minimum winter temperature is below –4°C and in the area where it is above 10 °C.

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(Total 10 marks)

   

Q11.          Parasites are organisms that live on or in host organisms. The populations of many organisms may be reduced by the effects of parasites.

Feather mites are small parasites found on the wing feathers of many birds. The mites feed on the oil that the birds produce. This oil keeps the feathers in good condition. Birds unable to oil their feathers properly use more energy in maintaining their body temperature. This results in less energy being available for other processes.

Scientists investigated the relationship between the numbers of feather mites and the breeding success of one species of bird, the great tit.

(a)     Use the information above to suggest how feather mites could affect breeding in great tits.

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(b)     The scientists located a large number of great tit nests. They sampled these at random.

For each nest they recorded

•        the total number of eggs laid

•        the number of chicks that hatched from the eggs

•        the number of chicks that survived to leave the nest

•        the total number of feather mites on the two parent birds.

(i)      Explain why the scientists sampled the nests at random.

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(ii)     The scientists calculated the percentage of each pair’s eggs from which chicks survived to leave the nest. They called this ‘breeding success per pair’.

The table shows some of the data that the scientists obtained.  

Total number of feather mites on both parent birds

Breeding success per pair

    0   86

    2 100

    5   64

  10   82

  14   70

  15   85

170   42

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Do these data support the hypothesis that the presence of feather mites reduces the ability of great tits to reproduce successfully? Give reasons for your answer.

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(c)     The scientists calculated a correlation coefficient for these data.

(i)      State a null hypothesis that would be appropriate for this investigation.

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(ii)     The correlation coefficient that they obtained had a negative value. What does a negative value indicate about these data?

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(d)     The oil that a great tit puts on its feathers is made in an oil gland at the base of the tail. The bird uses its beak to spread the oil over its feathers. This is called preening. Preening takes place in early morning and evening and empties the oil gland each time. After preening, the oil gland is considerably smaller.

At the same time that the scientists recorded the number of feather mites on each great tit, they also measured the size of the oil gland. The graph shows their results and includes the scientist’s line of best fit.

(i)      Describe the relationship between the number of feather mites present on each great tit and the size of the oil gland.

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(ii)     Explain how measuring the oil gland at the same time as counting the feather mites may have affected the reliability of the data.

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(e)     Feather mites eat pathogenic bacteria and fungi as well as oil. Explain how this may affect the breeding success of the birds.

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(Total 15 marks)

   

Q12. Malaria is a disease caused by a parasite. Scientists investigated the effect of malaria on competition between two species of Anolis lizard on a small Caribbean island. They sampled both populations by collecting lizards from a large number of sites on the island.

(a)     (i)       Explain the importance of collecting lizards from a large number of sites.

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(ii)     Describe one method the scientists could have used to ensure that the sites were chosen without bias.

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(iii)    The population number of both species of lizard varied at different times of the year. Suggest two reasons why.

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The scientists investigated the percentage of lizards of both species that were infected with malaria at different sites on the island. They collected samples of both lizards at intervals of 3 months for 1 year. They also recorded the elevation (height above sea level) of each site. Some of their results are shown in the table.

 

(b)     When analysing their results, the scientists used the percentage of lizards infected at each site, rather than the number of lizards infected. Explain why.

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  Site

Elevation of

collection site /

metres

Total number of

A. gingivinus collected in

one year

Percentage of A. gingivinus infected with

malaria

Total number of

A. wattsi collected in

one year

Percentage of A. wattsi

infected with malaria

  1 10 13 0 0 0

  2 80 30 0 0 0

  3 120 35 23 3 0

  4 200 40 30 7 0

  5 300 52 46 12 0

  6 315 35 31 13 1

  7 370 155 37 79 2

  8 414 124 44 68 4

(c)     A preliminary study suggested that malarial infections were more common at higher elevations. Use the information provided to evaluate this suggestion.

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(d)    (i)      As a result of this investigation, the scientists concluded that the presence of malaria provided a competitive advantage to A. wattsi. Use the information provided to explain how they reached this conclusion.

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(ii)     The malarial parasite of Anolis lizards destroys both red and white blood cells.Suggest how an increase in the percentage of A. gingivinus infected with malaria could result in A.wattsi having a competitive advantage.

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(iii)    The scientists carried out a statistical test to determine whether the correlation between the number of A. wattsi collected and the percentage of A. gingivinus infected was significant. They obtained a value for P of < 0.01.

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

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(Total 15 marks)

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M1.          (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

(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

+

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(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]

   

M2.          (a)    All / group of species / all/group of populations / all the organisms;

Accept equivalent terms for group. Answers which only refer to organisms must have idea of all the organisms not just a group of organisms Reject answers which include ‘environment’ or abiotic factors as part of the definition

1

(b)     (i)     7.2 - 8.4 (metres); Accept answer of 1.2

1

(ii)     1.      Food / prey / oxygen; Do not accept ‘resource’ for mark point 1 unless this is qualified as food / prey / oxygen

2.      Less / no competition; Reference to light and CO

2 as a resource negates mark point 2

Ignore intraspecific / interspecific for mark point 2 2

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(c)     1.      Increase in depth linked to decrease in temperature / decrease in depth linked to increase in temperature;

Accept increase or decrease in temperature is related to ‘higher depth’ or ‘lower depth’ due to ambiguity of these terms

2.      Correlation / relationship between temperature and fish distribution does not indicate a causal effect;

Ignore any reference to correlation unless it is clearly in context of temperature and fish distribution

3.      Overlap in ranges / different fish / species occupy same depth; Temperature does not determine fish distribution is not sufficient for idea of causal effect

4.      Other abiotic / biotic/named factor involved; Reject: ‘casual’ for mark point 2 Reject ‘other factors’ for mark point 4 unless further qualified

3 max [7]

 

M3. (a)    (Biological Agents)

Assume advantages are in context of correct type of control (chemical or biological) unless stated otherwise

1.     Only needs one application / reproduces;

2.     Specific;

3.     Keeps / maintains low population;

4.     Pests do not develop resistance; 4. Reject reference to immunity

5.     Can use less chemicals / reduces chemical residues / no bioaccumulation;

(Chemical pesticides)

6.     Acts quickly;

7.     Can apply to specific area;

8.     Kills all/most/greater variety of pests; 6 max

(b)     1.     Growth of algae / surface plants / algal bloom blocks light;

2.     Reduced / no photosynthesis so (submerged) plants die;

3.    Saprobiotic (microorganisms / bacteria); 3. Accept: Saprobiont/saprophyte/ saprotroph 3. Neutral: decomposer

4.     Aerobically respire / use oxygen in respiration;

5.     Less oxygen for fish to respire / aerobic organisms die; 5

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(c)     1.     Slaughtered when still growing / before maturity / while young so more          energy transferred to biomass / tissue;

Q 1-4 The principle here is one mark for identifying a relevant point and offering an explanation

2.     Fed on concentrate / controlled diet / so higher proportion of food absorbed / digested / assimilated / used for biomass / tissue / lower proportion lost in faeces;

2. Accept: named diets for controlled diet, eg high protein diet 2. Neutral: loss in excretion 2. Neutral: for growth

3.     Movement restricted so less heat / energy / respiratory loss;

4.     Heating / Kept warm / inside so less heat / energy / respiratory loss / maintain body temperature;

5.     Genetically selected / selective breeding (for high productivity); Neutral: reference to predators

4 max [15]

M4. (a)     1.   Provides a standard / benchmark;

2.   Can compare (different pesticides / chemicals);

3.   Does not kill all the tadpoles / organisms / population; 3. Accept ‘kills 50% of tadpoles’

2 max

(b)     1.      Only carried out on one species of toad / African toad / not carried          out on USA toads / tadpoles / species;

1. Accept not carried out on all species of toad 1. Accept carried out on different species 1. Do not accept one type of toad

2.   Only tested for 1 - 4 days / short term / not 16 days / long term;

3.   Did not look at effect of predator / predation; 3. Do not accept biotic factor on its own, there must be a reference to the predator

4.   Used various pesticides / may not have used malathion; 4. Accept’ did not use malathion’

3 max

(c)     (i)      1.      (See) effect of pesticide / malathion;

2.     Without predator / newts / stress / to compare / see effect with predator / newts / stress present (in experiment 2);

2

(ii)     1.    Tadpoles not killed / eaten;

2.    Newts are seen / detected; 2

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(d)     1.     Large surface area to volume ratio;

2.     Rapid / more diffusion / shorter diffusion pathway;

3.     Longer time exposure to pesticide / adults / toads live in and out of water / tadpoles remain / stay in water;

2 max

(e)     1.     Link between using less (pesticide) and cost / less effect on environment         / organisms;

1. Accept answers which link concentration (of pesticide) to being ‘cost effective’

2.      Pesticide / malathion diluted (in water);

3.      Concentrated due to evaporation;

4.      Concentrated in food chains / webs / tadpoles / habitat; 4. Accept (pesticide) ‘builds up’ in the environment / bioaccumulation / biomagnification

2 max

(f)      1.    Can see effect of other biotic factors / effect on other organisms;

2.    Can see effect of abiotic factors / named abiotic factor; 2

[15]

M5. (a)    0.8; 1

(b)    (i)      1.     Aerobic respiration; 1. Allow description e.g. respiration using oxygen 1. Accept ‘oxidative phosphorylation’

2.     Increase in uptake (of oxygen) with growth / reproduction / division of yeast cells;

3.     Glucose / nutrients / oxygen decreases / becomes limiting / cells die / ethanol/toxins form / heat produced / anaerobic respiration occurs; 3. Ignore any reference to time 3. Accept decrease in oxygen being linked to oxygen being ‘used up’ or equivalent

3

(ii)     1.     (Ethanol produced) by anaerobic respiration / from pyruvate         in anaerobic conditions;

1. ‘Fermentation’ is not enough on its own

2.     (Ethanol / anaerobic respiration) increases as oxygen (uptake / concentration) decreased;

3.     Decreases as glucose is used up / ethanol kills cells; 2 max

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(c)     1.     Oxygen uptake decreases / stopped;

2.     Oxygen is final (electron) acceptor / combines with electrons (and protons);

3.      Ethanol produced sooner / more ethanol produced; 3. Accept ethanol produced at any specified time before 16 hours

3 [9]

M6.          (a)     Krebs cycle/link reaction/pyruvate to acetylcoenzyme A;

Q Accept valid alternative for any of these steps. 1

(b)     (Respiratory reactions controlled by) enzymes;

Rate decreases as less kinetic energy/fewer collisions (between substrate and active site) fewer E-S complexes formed;

2

(c)     Requires hydrogen/electrons/is reduction;

Hydrogens from reduced NAD/reduced NAD reduces (pyruvic acid)/reduced NAD oxidised;

Information may be on diagram 2

(d)     Respiring anaerobically;

(Anaerobic respiration/respiration with nitrogen) less efficient/produces less ATP;

More anaerobic respiration/more glucose/substrate must be respired to produce same amount of ATP (so more carbon dioxide produced);

3 [8]

   

M7. (a)    (i)    So it / CO2 is not a limiting factor (on growth / photosynthesis);

Accept: CO2 is a limiting factor

1

(ii)     So any difference is due to iron (deficiency); Accept: iron is the variable

1

(iii)    Amount of triose phosphate / TP will be similar / same / low (at start); Accept: to allow triose phosphate to stabilise / become constant Reject: so all triose phosphate is used up Reject: so no triose phosphate

1

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(b)     1.     (Less) ATP produced; Accept: alternatives for reduced NADP ie NADP with hydrogen / s attached

2.     (Less) reduced NADP produced;

3.     ATP / reduced NADP produced during light-dependent reaction;

4.     (Less) GP to triose phosphate / TP; 4

(c)     1.     Less triose phosphate converted to RuBP; Accept: less triose phosphate so less RuBP

2.     CO2 combines with RuBP;

2 [9]

M8.          (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]

   

M9. (a)     1.     Expression / appearance / characteristic due to genetic constitution / genotype / allele(s);

1. Accept: named characteristic 1. Accept: homozygous / heterozygous / genes / DNA 1. Neutral: chromosomes

2.     (Expression / appearance / characteristic) due to environment; 2

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(b)     (i)      1.     3 and 4 and 9 / 11 / affected offspring; 1. Accept: 9 / 11 and their parents 1. Accept: unaffected parents have affected children

2.     Both 3 and 4 are carriers / heterozygous; 2. Accept: if 3 and 4 are unaffected all their children will be unaffected

OR

If dominant at least one of 3 and 4 would be affected; 2

(ii)     1.     11 is affected, 3 is not; 1 Accept: 3 / unaffected father / parents produce an affected daughter 1. Accept: 3 and 4 would only produce unaffected females

2.     3 / father of 11 does not have a recessive allele on his X chromosome / Xt;

2. Answers must be in context of alleles

OR

(If on X) 11 / affected female would not receive the recessive allele on X

chromosome / Xt from 3 / father;

Reject: recessive / dominant chromosomes

OR

(If on X) 3 / father (of 11) would pass on the dominant allele on his X chromosome / XT;

2

(c)     (i)      Answer in range of 5.8 − 6.2% = 3 marks;;; Answers in range of 0.058 - 0.062 = 2 marks

If incorrect answer, then 2 max of following points

1.     q2 / p2/ tt = 0.001 or 1 divided by 1000;

2.     p / q / T = 0.968 − 0.97;

3.     Understanding that heterozygous = 2pq; 3. This can be shown mathematically ie 2 × two different numbers 3. Accept: answer provided attempts to calculate 2pq

3 max

(ii)     Affected individuals (usually) do not reproduce / die during childhood / do not pass on allele / genetic screening;

1 [10]

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M10.         (a)     1. Occurs in an unchanging environment; 1

+ 2. (Initial range of values in which) mean is best adapted; 3. Selection against extremes / selection for the mean; 4. Mean/median/mode unaltered 5. Range/S.D is reduced; 6. Repeated over many generations; 7. Increasing proportion of populations becomes well adapted     to environment;

max 4

(b)     1. All plants are acyanogenic below –4 °C and (most) cyanogenic     above +10 °C; 2. Cyanogenic plants’ cells freeze below –4°; 3. Releasing cyanide (into their own tissues); 4. Damaging/killing plants / disrupting metabolism; 5. Selective advantage not to produce cyanide at –4 °C; 6. Slugs present at higher temperatures / not usually present/inactive at     lower temperatures; 7. Cyanide production kills/deters slugs; 8. Advantage only at higher temperatures;

max 5 [10]

   

M11.          (a)     1. Breeding less successful;

2. Feathers in poor condition;

3. Less energy for breeding/reproduction / stated aspect of reproduction; 1. Reject cannot breed. 2. Ignore "wings damaged".

2 max

(b)     (i)      1. Avoids bias;

2. Data representative/choice of nest not influencing results;

3. Allows use of statistical tests/named statistical test; 2 max

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(ii)     Accept general statements or statements based on data that make the required points.

1. Correct statement about range of 0 – 15;

2. Correct statement about 0;

3. Correct statement about 170;

4. Correct statement about gap between 15 and 170; 1. e.g. No pattern/no correlation between 0 and 15. 2. e.g. Birds with no feather mites did not have (the) high(est) breeding success / 86%. 3. e.g. Highest number of feather mites linked to lowest breeding success. 4. e.g. No data between 15 and 170.

3 max

(c)     (i)      There is no correlation between the number of feather mites and breeding success / the number of feather mites does not affect breeding success;

These specific variables must be stated. Reject difference between feather mite and breeding success.

1

(ii)     Breeding success decreases as feather mites increases/ negative correlation between feather mites and breeding success;

Accept reproductive or breeding success. 1

(d)     (i)      1. The larger the size of the oil gland the larger the number of feather mites;

2. Positive correlation;

3. (Wide) scatter of points / points not on line; 3. Accept any answer that conveys the idea of a wide spread. Ignore any reference to anomalies.

2max

(ii)     No mark for effect on reliability, marks are for explanation.

1. Oil gland size/number of mites could vary;

2. At different times of the day/due to preening; Ignore responses that state oil gland affects numbers of mites. Allow preening affects mite numbers/size of oil gland.

2

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(e)     1. Improve health of birds/reduces disease/reduces harm;

2. Healthier birds may find more food for young/do not pass on disease/ have     greater specified aspect of breeding success;

1. Ignore death of birds. 2. Specified aspect can include longer breeding life.

2 max [15]

   

M12. (a)     (i)      Reliable / representative / for statistical tests;

Accept: identify anomalies Neutral: accurate/valid/bias

1

(ii)     1.     Find coordinates (on a grid) / split area into         squares / number the sites;

1. Ignore references to tape measures, metre rulers etc

2.     Method of generating / finding random numbers eg calculator / computer / random number generator / random numbers table; 2. Accept: numbers out of a hat / use of dice

2

(iii)    1.     Breeding (of lizards); Neutral: weather / climate / hurricanes / hibernation / migration / emigration / immigration

2.     Food source / prey;

3.    Predator;

4.     Variation in malarial infection;

5.     Temperature variation;

6.     Availability of water eg drought/‘rainy season’ 2 max

(b)     1.     Number in sample varies;

2.     Allow a (valid) comparison; 2

(c)     1.     (Overall) positive correlation (for either / both species); Neutral: only one study / no repeats

2.     Reference to (site) 5 / 300 metres;

3.     Limited results for A. wattsi / small sample / number / percentage infected for A. wattsi;.

2 max

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(d)     (i)      1.     Fewer A.wattsi infected / more A. gingivinus infected;

2.     Higher number of A.wattsi present when higher percentage / number of A.gingivinus infected;

3.     No A.wattsi present when A.gingivinus has zero infection; 2 max

(ii)     1.     Reduced immunity / increased susceptibility to disease; 1. Accept: idea that energy / resources are used to combat malaria

2.     Reduced oxygen transport / uptake / respiration / reduced activity / movement;

2

(iii)    1.     There is a probability of less than 1% / 0.01; 1. Reject: probability is / equal to 1% / 0.01; 1. Reject 0.01% / 5% / 0.05 / 0.05%

2.     That result(s) / correlation / it is due to chance; 2. Allow correct interpretation using above (incorrect) figures eg there is a probability of less than 5% that the results are due to chance =1 mark

OR

3.     There is a probability of more than 99% / 0.99;

4. That result(s) / correlation / it is not due to chance; Note: there is a probability of more than 5% that the results are due to chance =0 marks 3. Reject: probability is / equal to 99%/0.99; 3. Reject 0.99% / 95% / 0.95 / 0.95% 4. Allow correct interpretation of above figures ie 0.99% / 95% / 0.95 / 0.95% but reject if less than

2 [15]

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E1.          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.

   

E2.          (a)     Four out of five students were able to provide a suitable definition of the ecological term community. Incorrect responses were often definitions relating to a population or ecosystem.

(b)     (i)      Almost two thirds of students correctly provided the range of depths, 7.2 to 8.4, at which all three species of fish may be found living together. The most common incorrect response was 6.8 to 8.4.

(ii)     The vast majority of students obtained both marks for this question often by referring to competition for food. Students gaining one mark usually described competition for an unspecified ‘resource’.

(c)     Most students gained one mark for a description of the decreasing temperature with increasing depth of water. Approximately half of these students then gained another mark for describing either the overlap of the ranges of the fish or the idea that another named factor could be involved. Very few students mentioned both of these ideas to obtain 3 marks. Some students merely described the depth ranges and temperatures at which each of the three species of fish were found or referred to the unknown sample size or lack of statistical tests. Although a significant number of students referred to correlation not indicating causation, this was more often than not described in an incorrect context.

   

E3. (a)     Only ten percent of students obtained more than four out of the six marks available. Answers often gave the disadvantages of chemical pesticides and biological agents without making the advantages of using an integrated system clear. All marking points were seen in responses, but usually only 2 or 3 of them in any one response. The most common marking points referred to the specificity of biological agents, its single application and the reduced need to apply chemical pesticides. The fact that chemical pesticides act quickly was also appreciated by many students. A common error in weaker responses was to confuse pesticides with fertilisers and then to provide a detailed account of eutrophication. On a positive note, the common confusion between resistance and immunity was less apparent than on previous papers. A larger than expected number of students gained no marks for this question.

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(b)     Not surprisingly this question produced a lot of good answers but still discriminated well despite forty percent of students scoring four or more marks out of the five available. Weaker responses lacked the appropriate level of scientific terminology, omitted essential details or confused ideas. Most students referred to an algal bloom and its effect on penetration of light. However, some students omitted any reference to photosynthesis, or related a reduced oxygen concentration solely to the activity of plants. Some students referred to fish dying due to lack of food with no reference to oxygen or respiration. The best responses were often clear and concise and read as the mark scheme. These answers referred to saprobiotic microorganisms rather than simply ‘decomposers’ and clearly related the death of fish to a decrease in oxygen for respiration.

(c)     Almost eighty percent of students obtained at least half the marks on this question. However, relatively few obtained maximum marks despite all the marking points regularly occurring on different scripts. The two most commonly seen marking points were those related to less respiratory losses due to animals being kept inside / warm and having restricted movement. References to selective breeding were also fairly frequent. Students often referred to animals being fed controlled diets but did not always explain clearly how this increased net productivity. A common explanation, which was not credited, was to state that it increased ‘growth’. Similarly, when students referred to animals being slaughtered when still growing, explanations were often inadequate. Better students had no such problems and provided verbatim the mark scheme answers.

E4. (a)     Very few students were able to suggest two advantages of using the LC50 to determine the effect of a pesticide. Correct responses usually referred to not killing all the tadpoles, or for comparing different pesticides. Incorrect responses were often rather confusing and focused on statistical tests, finding the most cost-effective concentration or on reliability.

(b)     This question proved to be relatively straightforward for most students, who often gained at least two marks. They appreciated that previous studies have been carried out mainly on one species of toad and for a relatively short time period. However, many students did not make the link between a biotic factor and a predator. In fact, there was a significant minority of students who described a biotic factor as being temperature or pH.

(c)     (i)      The majority of students gained at least one mark, usually for explaining that the scientists could see the effect without the newts being present. Less than half these students gained a second mark for mentioning that the scientists would see the effect of the pesticide on the tadpoles. Students failing to gain any marks often simply stated that experiment 1 was a control with no further qualification.

(ii)     Almost ninety five percent of students gained at least one mark, usually for the idea that the newts are visible to the tadpoles. More than half these students gained a second mark for stating that the barrier prevented the tadpoles being eaten by the newts.

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(d)     Most students had difficulty obtaining both marks on this question. Many students gained a mark for explaining that the tadpoles would be exposed to the pesticide for a longer time period, as they stay in the water. However, relatively few students linked the smaller size of tadpoles to a larger surface area to volume ratio and even fewer students mentioned a shorter diffusion pathway.

(e)     This proved to be a very demanding question, with two-thirds of students scoring zero. Most students failed to understand why a range of concentrations of the pesticide was used. Many answers only described finding the most effective concentration to use, to see the effect on other organisms or to see if the tadpoles developed resistance; to get a reliable mean, to identify anomalies or to use a statistical test also featured. When marks were awarded, it was usually for the idea of the pesticide building up in the environment, being diluted or for the idea of finding the lowest effective concentration that causes least harm to the environment.

(f)     Weaker answers were limited to responses such as ‘to obtain valid results’, ‘see effects in the real world’, or suggested that these investigations would be cheaper. However, better answers showed appreciation that these investigations would enable scientists to see the effects of abiotic factors on the pesticide and / or the effect of the pesticide on other organisms.

E5. (a)     Approximately half the students correctly calculated the rate of oxygen uptake as 0.8 arbitrary units.

(b)     (i)      Most students gained two marks for this question. They linked the uptake in oxygen to aerobic respiration and then provided one of a range of correct responses to explain the decrease in uptake including: oxygen / glucose becoming limiting, ethanol causing death of cells and, most commonly, anaerobic respiration occurring. Very few students gained a third mark by linking the increase in uptake of oxygen to the growth / reproduction of the yeast cells. Weaker answers linked ethanol production to oxygen uptake, or suggested that the yeast would carry out photosynthesis.

(ii)     The vast majority of students appreciated that the ethanol was produced during anaerobic respiration. However, only half these students linked the increase in ethanol production to the fall in oxygen uptake. Relatively few students referred to the decrease in ethanol production as the glucose was used up.

(c)     Overall this question was well answered by students. The role of oxygen in the ETC was well known and there were some very clear answers that gained all three marks. The decreased uptake of oxygen and the earlier and greater production of ethanol were often described. Occasionally, the sodium azide was believed to have no effect on ethanol production. NADP was sometimes included rather than NAD in descriptions of the ETC and there were disappointing references to energy being produced. Some confused responses suggested that oxygen was needed for ethanol production and without oxygen being used, no ethanol would be produced. Weaker answers described yeast as a plant and referred to the Calvin cycle.

E6.          The weighting of the assessment objectives limits the number of marks that can be awarded for recall of basic factual information and this question sought to test understanding of the principles underpinning respiration. It became clear in marking the scripts that although many candidates had considerable knowledge of the appropriate technical language, they lacked fundamental understanding.

(a)     Although there were occasional references to glycolysis, most candidates correctly named either the Krebs cycle or the link reaction.

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(b)     Although most candidates recognised the role of enzymes, the examiners were left with the impression that many of those who made the basic error of linking decreasing temperature to enzyme denaturation were responding with prepared answers. A disturbing number of candidates revealed here, and elsewhere in the question, confusion between respiration and photosynthesis.

(c)     Apart from predictable and confusion between oxidation and reduction, and between NAD and NADP this question was answered well.

(d)     Although this part of the question was targeted at those candidates likely to be awarded higher grades, it was disappointing to note that relatively few were able to make use of the information provided, that the apple slices were transferred to anaerobic conditions in pure nitrogen gas. Many wrote about nitrogen fixation, bacterial decomposition or the need for nitrogen in photosynthesis. Better candidates generally identified respiration as anaerobic even if they failed to discuss the relative inefficiency of the process.

   

E7. (a)     (i)      Three out of four students gained this mark by stating that carbon dioxide is a limiting factor on photosynthesis. Inadequate responses often simply stated that carbon dioxide was needed for photosynthesis.

(ii)     Twenty percent of students gained this mark by indicating that any difference in growth during the experiment would be due to iron deficiency. There was a large variety of incorrect responses but the majority referred to a ‘fair test’, ‘similar growth’ or ‘same level of TP’.

(iii)    Only the better responses included the realisation that the period in the dark was to ensure that the levels of triose phosphate were similar / low in both groups of plants at the start of the experiment. Many who did mention triose phosphate thought that it would all have been used up or converted to glucose / RuBP, ignoring the evidence from the graph. Weaker responses only referred to preventing the light-dependent reaction, or just preventing photosynthesis.

(b)     This question proved to be a very effective discriminator and produced a good range of marks. Most students gained a mark for ATP being produced during electron transport. Although many students then gained a mark for the production of reduced NADP, a significant minority incorrectly referred to NAD. The decrease in production of triose phosphate was not always linked to a decrease in glycerate-3-phosphate. Only better responses referred to ATP and reduced NADP being produced during the light-dependent reaction. Despite the stem of the question stating ‘knowledge of photosynthesis’ a number of students described electron transport during respiration.

(c)     This question caused some difficulty for students, with almost forty percent gaining no marks. It was surprising to find a number of students referring to respiration in their answers and confusing the Calvin cycle with the Krebs cycle. Some students with a better understanding failed to gain credit as they limited their explanation to the Calvin cycle being reduced. Students gaining one mark often referred to the role of RuBP in combining with carbon dioxide. Only the best responses linked iron deficiency to a reduction in the amount of triose phosphate and therefore less RuBP.

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E8.          (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.

   

E9. (a)     The phenotype as the expression of the genotype was well known, with the vast majority of students gaining the first marking point. Less than a quarter of students mentioned the effect of the environment on the phenotype. Answers that failed to gain a mark just described the phenotype as the visible characteristics or appearance of an organism.

(b)     (i)      Over a third of students scored zero on this question. Most of these students suggested that the phenotypes of 1, 2, and 4 / 5 proved that the allele for Tay-Sachs disease is recessive. There was also a common misconception that individual 3 was the son of 1 and 2. Approximately half the students who correctly referred to individuals 3, 4 and 9 / 11, provided a suitable explanation to gain a second mark. Unfortunately, the use of poor terminology prevented more students from gaining both marks; for example, ‘do not carry the gene’.

(ii)     This proved to be very problematic with less than twenty percent of students scoring above zero. Many students had no idea how to prove that the allele for Tay-Sachs was not on the X-chromosome, often identifying the wrong individuals or suggesting that only males should be affected. Some students gained the first mark for correctly selecting individuals 11 and 3, but only a minority of students were able to explain how this proved that the allele was not on the X-chromosome. The allele was often thought to be on the Y-chromosome. The fathers were also thought to pass their X-chromosome to their sons. Some answers were very general and failed to mention any numbered individuals.

(c)     (i)      This proved to be quite an effective discriminator despite fifty percent of students obtaining all three marks. Students obtaining a single mark often did so by attempting

to calculate 2pq. Although many students gained credit for showing that q2 = 0.001, a significant number incorrectly indicated that q = 0.001. It was disappointing to find incorrect answers which showed no ‘working’ as some of the calculation might have been along the right lines. The required use of decimals certainly proved troublesome for some students. A small minority of students simply provided the Hardy-Weinberg equation but made no attempt to use it.

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(ii)     It was surprising to find that almost fifty percent of students did not gain this mark. Most correct responses noted the information provided in the stem of part (b), ie ‘sufferers of this disease often die during childhood’. Incorrect responses related to more general reasons why the Hardy-Weinberg equation might not apply, such as emigration, immigration, lack of random mating etc. Random fertilisation and mutation were often seen as incorrect responses. Natural selection, selection against individuals with Tay-Sachs, stabilising selection and directional selection were also suggested.

E10.          Much of the material in this question was unfamiliar and required careful analysis to score highly. Even so, candidates across the ability range scored highly on the question, although it was mainly able candidates who did so.

(a)     Many answers to this question, which demanded only straight recall, were disappointing. Most candidates did not get beyond saying that stabilising selection acts against the extremes and results in a smaller range or standard deviation. They often cited the example of human birth mass. Better candidates were able to explain that this occurred because those individuals showing the mean value were best adapted to the environment and that, as a consequence, the mean would not change as the process repeated itself generation after generation. However, only a few of those took notice of the question and explained that stabilising selection occurs in an unchanging or stable environment.

(b)     Candidates who made the link between temperature, the activity of slugs and the proportions of cyanogenic and acyanogenic clover plants usually went on to score well by explaining the selective advantage of producing cyanide when slugs are active and not producing cyanide when temperatures are so low that the cyanide would be released into the plant’s own tissues. Those who did not make this link often pursued an inappropriate line of reasoning concerning the effect of temperature on enzyme kinetics and developed highly teleological answers.

   

E11.          (a)     Candidates frequently failed to gain credit because they wrote that the mites affected breeding (as stated in the question) rather than reduced breeding success, before they went on to explain why.

(b)     The response 'to eliminate bias' gained most candidates one mark for their answer to part (i), but fewer appreciated the importance of being able to apply a statistical test to the data collected. In part (ii), many candidates showed that they understood the significance of variation in breeding success but did not look critically at the independent variable so did not comment on the lack of data between 15 and 170 mites per pair of parent birds. Weaker candidates tried to answer the whole question in a single sentence rather than breaking their answers into separate parts.

(c)     In part (i), candidates seemed to find it difficult to go beyond the general answer of 'there is no difference' and relate their null hypothesis to the investigation in question. A common error was to state that, 'There is no difference between the mites and the breeding success of the birds'. Most incorrect responses to part (ii) simply restated that the correlation was negative, or that one value went up as the other went down. Marks were only gained by answers that related to the data in the question, stating clearly that breeding success went down as the number of mites went up.

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(d)     Many candidates made reference in part (i) to anomalous data, rather than describing the wide scatter or points. Others failed to gain marks because they did not refer to the number of mites and the size of the oil gland in their answers. In part (ii), some candidates only considered the practical details here, rather than using the information in the stem. They were required to explain that preening only took place at certain times of day and that after preening the gland was significantly smaller, causing the measurement to be lower than it would have been earlier in the day.

(e)     Most candidates realised that pathogenic organisms were harmful, but few made the second step and explained that a reduction in pathogens would allow the parent birds to have greater breeding success because, for example, they had not passed disease to their offspring.

   

E12. (a)     (i)      Almost eighty percent of students obtained this mark, usually by stating that this method helps to make the samples ‘representative’ or ‘reliable’.

(ii)     The vast majority of students gained at least one mark for referring to using coordinates or dividing an area into squares. Most of these students gained a second mark for describing a method of generating random numbers. Some students incorrectly referred to the mark, release and recapture method.

(iii)    Almost fifty percent of students gained two marks for providing two valid reasons for changes in the population of lizards at different times of the year. Breeding, predators, prey and temperature changes were common correct responses. Responses such as ‘weather’ or ‘climate’ were not credited and limited most of the other students to one mark.

(b)     This was generally answered well, with many students gaining both marks. The most common scoring point was the idea of allowing a valid comparison. Some students failed to note that the numbers in the samples vary and obtained one mark. Incorrect responses often referred to reliability.

(c)     Considering the data provided were not particularly complex, it was disappointing to note that forty percent of students scored zero for this question. The most common scoring mark point was for referring to the results at site 5 for A. gingivinus. Relatively few students gained a second mark by referring to a positive correlation or to the limited results for A. wattsi.

(d)     (i)      The vast majority of students gained one mark for stating that more A. gingivinus than A. wattsi are infected. Many students then tried to explain why this might be the case (more die so less competition for food etc), rather than trying to give further explanations of how they reached the conclusion. As a result, very few students obtained two marks.

(ii)     A significant minority of students did not use the information provided on the destruction of blood cells and gained zero marks. Students who did use this information often referred to the increased susceptibility to disease due to destruction of white blood cells. There were slightly fewer marks gained for the effects caused by a reduction in red blood cells.

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(iii)    Less than fifty percent of students obtained both marks. Although students understood that the P value represented probability, the less than symbol < was often interpreted as ‘is’ or ‘equal to’. Other errors included referring to 0.01% rather than 1% probability or the correlation not being due to chance (with 0.01/1%). Some students correctly used a probability of more than 99% / 0.99 that the correlation is not due to chance.

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