b) Identify the optimum mass at birth for survival. (1)



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1. Researchers carried out a study on 3760 children born in a London hospital over a period of 12 years. Data was collected on the childrens’ mass at birth and their mortality rate. The purpose of the study was to determine how natural selection acts on mass at birth. The chart shows the frequency of babies of each mass at birth. The line superimposed on the bar chart indicates the percentage mortality rate (the children that did not survive for more than 4 weeks).

[Source: W H Dowderswell, (1984) Evolution, A Modern Synthesis, Heinemann Educational Books, page 101]

(a) Identify the mode value for mass at birth.

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(1)

(b) Identify the optimum mass at birth for survival.

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(1)

(c) Outline the relationship between mass at birth and mortality.

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(2)

(d) Explain how this data supports natural selection.

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(2)

(e) Suggest one environmental factor that could cause a low mass at birth.

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(1)

(Total 7 marks)


2. The mosquito (Wyeomyia smithii) uses daylength as a guide to either continue development of its larvae or to begin hibernation. This response to daylength is genetically controlled. Longer daylengths maintain development whereas shorter daylengths induce hibernation. In the northern regions of the northern hemisphere, even though daylengths are longer, winter arrives earlier than in regions closer to the equator. The following data is from an experiment to determine if W. smithii has adapted to later onsets of winter as a consequence of global warming. In 1972 and 1996, larvae were collected at various locations in the United States at latitudes 30–50° North. The larvae were examined to determine what daylength induced hibernation. Each circle on the following graph represents one larval population.

[Source: Bradshaw and Holzapfel, Proceedings of the National Academy of Sciences of USA, (2001),


98 (25), pages 14509–14511]
(a) Outline the relationship between daylength and latitude for the larval populations in 1972.

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(1)

(b) Compare the data of 1972 with 1996.

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(2)

(c) Explain how the data illustrates an evolutionary response to a longer growing season due to a later onset of winter.

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(2)

(Total 5 marks)

3. Plants called epiphytes grow above the ground on the surface of other plants such as trees. Epiphytes play an important role in rainforests because they can absorb vast amounts of precipitation water (rain and fog), retain minerals effectively and contribute enormous amounts of humus.

An investigation was carried out in Ecuador to determine the distribution and abundance of epiphytes in lowland and mountain rainforests. Branch cover was measured for different branch diameters and different branch angles. (The larger the branch diameter, the closer it was to the trunk of the tree.) The results are shown below.



[Source: Freiberg and Freiberg, Journal of Tropical Ecology, (2000), 16, pages 673–688]

(a) Outline the percentage of branch cover on horizontal branches of trees in mountain rainforests.

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(1)

(b) Analyse how the branch angle affects the percentage of branch cover in mountain rainforests.

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(2)
(c) Compare the percentage of branch cover between mountain rainforests and lowland rainforests for inclined branches.

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(2)

(d) Suggest a reason for the overall difference in branch cover by epiphytes in mountain rainforests and lowland rainforests.

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(1)

(Total 6 marks)


4. Seed dispersal is important in the migration of plants from one area to another area. Plants have evolved many methods, both physical and biological, by which to disperse their seeds.

50 maple seeds, which are wind dispersed, were dropped one at a time from two different heights, 0.54 m and 10.8 m respectively. The histograms below show the distribution of the distance the maple seeds travelled.



[Source: student experiment, Guralnick]


(a) For each height, identify the distance travelled by the greatest number of seeds.

(i) Height = 0.54 m: ...............................................................................................

(ii) Height = 10.8 m: ...............................................................................................

(1)

(b) State the effect of height on seed dispersal.

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(1)

(c) Suggest two reasons for the effect of the drop height on the distance travelled by the seeds.

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(2)
The following graphs show the rate and timing of seed release from different species of grass in the same area during the summer.

[Source: J L Harper, Population Biology of Plants, Academic Press (Harcourt Brace Jovanovich) 1997, page 57]

(d) Identify the grass species which produces the most seeds in this area.

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(1)

(e) Identify the grass species which produces the most seeds in June.

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(1)
(f) Compare seed production for all species relative to the timing of their release.

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(3)

(g) Suggest two benefits for these plants in the timing of seed release.

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(2)

Biological seed dispersal is usually dependent on the nutritional content of the seed or fruit. The following table gives the nutritional content for fruits of different species in temperate and tropical climates.









Percentage by Dry

Weight




Common Name (genus)

Protein Lipid

Carbohydrate

Dispersal Agents







Temperate







Cranberry (Vaccinium)

3

6

89

Birds

Hawthorn (Crataegus)

2

2

73

Birds

Pin cherry (Prunus)

8

3

84

Birds

Pokeberry (Phytolacca)

14

2

68

Birds

Strawberry (Fragaria)

6

4

88

Birds







Tropical







Bird palm (Chamaedorea)

14

16

55

Birds

Fig (Ficus)

7

4

79

Bats

Mistletoe (Viscum)

6

53

38

Birds

Monkey fruit (Tetragastris)

1

4

94

Monkeys

Wild nutmeg (Virola)

2

63

9

Birds

[Source: H Howe and L Westley, Ecological Relationship of Plants and Animals,
Oxford University Press 1988, page 121]

(h) Compare tropical fruits to temperate fruits in relation to the mean values for lipid, carbohydrate and protein content.

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(2)
(i) Explain which fruit would have the highest energy content.

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(2)

(j) Suggest one advantage and one disadvantage of dispersal of seeds by animals.

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(2)

(Total 17 marks)

5. Osteoporosis is a major health problem for many post-menopausal women. As the ovaries reduce their secretion of estrogen, calcium is gradually lost from bones, weakening them and increasing the chance of fractures. To test whether diet influences the rate of calcium loss, ovaries were removed from groups of female rats and the rats were then either fed a control diet or the same diet with one gram of a supplementary food per day. The rate at which the rats excreted calcium was measured. The ratio of calcium loss between the control rats and the rats that were given a supplementary food was calculated. .

The results are shown in the graph below.



[Source: Muhlbauer and Li, Nature, 1999, 401, pages 343–344]

(a) (i) Identify which supplementary food was most effective in reducing calcium loss.

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(1)

(ii) Identify which supplementary food was least effective in reducing calcium loss.

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(1)
(b) Among the ten foods shown in the graph, seven are plant products (vegetables) and three are animal products. Discuss whether the plant or the animal products were more effective at reducing calcium loss.

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(3)

(c) Suggest a trial, based on the results shown in the graph, that could be done to try to reduce osteoporosis in humans.

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(3)

(Total 8 marks)


1. (a) any value from 3.25 to 3.49 kg
(could be U - 1) (range not required) 1

(b) any value from 3.50 to 3.74 kg


(could be U - 1) (range not required) 1

(c) initially as birth mass increases /


up to 3.5 kg, survival increases /
mortality decreases;
then, as birth mass further increases /
beyond 3.5 kg, survival decreases /
mortality increases;
further from mode the higher the mortality /
highest survival / lowest mortality
nearest to mode value; 2 max

(d) birth mass shows variation;


selection against very low / very high birth weights; 2 max

(e) mother’s diet / smoking / pollution /


mother’s alcohol consumption / mother’s health
/ mother’s blood pressure (not inherited) / disease (not inherited) 1
Only one answer is acceptable.
If several answers are given, mark the first one.


[7]


2. (a) as the latitude increases so the length of day
needed to induce hibernation increases;
there is a direct correlation between
daylength and latitude; 1 max

(b) the shortest daylength to induce hibernation is the same


at the lowest latitude in 1972 and 1996;
at higher latitudes, the daylength to induce hibernation is
less in 1996 than 1972;
for the same daylength, the latitude at which hibernation occurs
has moved further north in 1996 compared to 1972;
at lower latitudes the change in daylength is less
than at higher latitudes; 2 max
(c) colder areas / higher latitudes become warmer /
resemble lower latitudes (with shorter daylength);
warmer weather delays organisms from beginning
hibernation / mosquitos stay active longer into winter;
(the graphs show) that organisms at higher latitudes have
adapted to a shorter daylength (to begin hibernation); 2 max

[5]


3. (a) thicker branches have more cover /
branches near the tree have more cover;
thinner branches have less cover /
no change until the branch is 17.9–1 cm; 1 max

(b) with higher angles there is usually less coverage;


there is less change between 0–30° and 31–60°
than between 31–60° and 61–90°;
the coverage for horizontal and inclined
branches is almost the same;
the ends of branches show least coverage; 2 max

(c) the overall patterns are the same;


for branch diameters of 40–9 cm there is 100% coverage
in the mountains compared to 40% in the lowland forest;
for thinner branches (8.9–1 cm) the percentage
cover in lowland forests is almost zero
whereas mountain branches have 40% cover; 2 max

(d) mountain forests are often covered in mist or clouds


so there is more moisture available;
more precipitation in mountain forests;
lowland forests are warmer and plants dry out more easily;
more animal activity / grazers in lowland forest; 1 max

[6]


4. (a) height 0.54 m: 60–79 cm / 0.60–0.79 m (from the plant)
and height 10.8 m: 0–2.9 m
(from the plant); 1
Units needed for both parts of the answer.

(b) the greater the height from which the seed fell,


the further it travelled from the parent plant 1
(c) at the greater height:
seed can catch the wind to travel further / updrafts /
more wind at greater height;
farther to the ground and does not travel straight down /
more time to be blown before hitting the ground;

at lower height:


seed can fall straight down;
seed can hit downdraft and fall faster; 2 max
Any point must explain the difference in distance
travelled from the two heights.

(d) Agrostis stolonifera 1

(e) Poa trivialis 1

(f) Poa produces seed earliest in the summer / June;


Holcus produces most seed in July;
Agrostis and Festuca produce seed in (late July to) August;
Holcus and Poa have a peak time of seed fall /
short period of seed fall;
Agrostis and Festuca may continue to increase in
seed production to September; 3 max
Accept any of these points made conversely as an alternative.

(g) Award [1] each for any two of the following.


to avoid predation /
disperse at times when other species are dispersing their seeds;
to avoid competition;
late in the year to allow seeds to germinate over winter /
better germination conditions;
better dispersal conditions / more wind / animals for dispersal;
photoperiod – required day length for flowering;
more energy stored at the end of the summer for seed production;
more light / warmth / better conditions for seedling photosynthesis /
growth; 2 max

(h) Award [1] each for any two of the following.


tropical fruits have higher lipid content than temperate fruits;
temperate fruits (80%) have greater carbohydrate
content than tropical fruits (55%);
protein levels are similar in both groups of fruits /
slightly higher in temperate fruits
than tropical fruits;
(must make it clear that the difference is slight) 2 max
(i) mistletoe;
high proportion of lipid and carbohydrate
(lipid has approximately twice the energy content
of protein and carbohydrate); 2

(j) Award [1] for advantage and [1] for disadvantage.

animal dispersal advantage:
travel further / digestion cracks seed coat for better germination /
deposited in feces with organic matter /
better in areas with little wind;

animal dispersal disadvantage:


predation / seeds eaten / deposited in poor environment /
buried too deep / buried too shallow (if deposited with feces) /
animal might become extinct / scarce; 2 max

[17]


5. (a) (i) parsley; 1

(ii) egg; 1

If the answers to A1 (a) (i) and (ii) are correct, but inverted, ie (i) egg and (ii) parsley, award [0], but apply Error Carried Forward (ECF) to marking points (b) and (c). Write ECF beside the answer and use the alternative marking points given below.

(b) plant products more effective overall / animal products least effective;


the three / four / five most effective foods are all plant products / ratios given of 3 plants; but soy-beans is an exception / similar to egg; potato and skimmed milk are similar in effectiveness;

With ECF:

plant products least effective overall / animal products more effective;


the three / four / five least effective foods are all plant products / ratios
given of 3 plants;
but soy-beans is an exception / similar to egg;
potato and skimmed milk are similar in effectiveness; 3 max

(c) give supplements of parsley / garlic / onion to a group of people;


use women after the menopause / with osteoporosis;
use women with ovaries removed;
have another control group of similar people who are not given the supplement;
measure changes in bone density during the trial period / measure calcium loss;
With ECF:

give supplements of egg / meat / soy-bean to a group of people;


use women after the menopause / with osteoporosis;
use women with ovaries removed;
have another control group of similar people who are not given
the supplement;
measure changes in bone density during the trial period / measure
calcium loss; 3 max

[8]





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