17. Selection and evolution
A section of Biology, 9700
Listing 10 of 104 questions
[T • There are over 40 Galapagos Islands including the small and isolated island named Daphne Major. • Only two species of Darwin finches are found on this island. • Studies were made every year from 1970 to 1989 on the beak size of the island’s population of ground finch, Geospiza fortis, by measuring the beak length of every bird (). Larger finches with larger beaks are better at opening large seeds. From 1976 to 1978 there was a drought and only 15% of the ground finches survived and these did not breed during drought years. • All finches were reduced in number. The most conspicuous feature of the survivors of the drought years was their large beak size. • The main environmental consequences of drought is the decline in food supply, mainly seeds. During normal years, many grasses and herbs produce an abundance of small seeds. A few other plants produce a much smaller number of large seeds which are not normally eaten. Describe how environmental factors appear to have acted, during drought years, on the beak size of finches as an evolutionary force of natural selection. Finches with small beaks were found to be smaller than finches with larger beaks. Explain the stabilizing force of natural selection on the beak size and size of birds in normal years. Use Outline the mechanisms that may have let natural selection lead to the evolution of the thirteen species of Darwin finches now found on the Galapagos Islands. [Total : 8]
9700_w05_qp_4
THEORY
2005
Paper 4, Variant 0
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Use Asellus aquaticus is a small freshwater crustacean. 200 A. aquaticus were released into a pond where there had previously been none. The pond was favourable for their growth and reproduction. Describe and explain the expected changes in the population size of A. aquaticus over the following few months. In order for natural selection to occur a population must show phenotypic variation. Explain why variation is important in natural selection.
9700_w09_qp_41
THEORY
2009
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Use A recent study of the house mouse, Mus musculus, on the island of Madeira resulted in the following observations. ● There are six distinct populations. ● The mice are associated with human settlements. ● The populations are located in different valleys separated by steep mountains. ● Each population has a different diploid number of chromosomes. As a result of these observations it has been suggested that speciation is taking place. is a map of Madeira showing the distribution of the six populations. mountains Using the information in , state the likely isolating mechanism and the type of speciation taking place. isolating mechanism type of speciation Examiner’s Use Explain how speciation is occurring in the house mouse populations of Madeira.
9700_w09_qp_42
THEORY
2009
Paper 4, Variant 2
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From 1975 to 1977 one of the Galapagos Islands, Daphne Major, experienced a severe drought. A ground finch, Geospiza fortis, feeds on seeds on Daphne Major. shows the ground finch, G. fortis. • One of the few plants that survived the drought produced large seeds inside tough fruits. • Many G. fortis died during the drought and the population declined from 1400 in 1975 to 190 in 1977. • The mean beak depth of those G. fortis that died was 10.68 mm and the mean beak depth of those that survived was 11.07 mm. • The mean beak depth of G. fortis before the drought was 10.86 mm. Calculate the percentage decrease in population size between the years 1975 and 1977. Show all the steps in your calculation and give your answer to the nearest whole number. Answer % Suggest why some G. fortis were able to survive the drought while others died. Natural selection was taking place on Daphne Major. State the type of natural selection operating on G. fortis during the drought and explain your answer. type of natural selection explanation
9700_w10_qp_43
THEORY
2010
Paper 4, Variant 3
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The following passage is a summary of the main principles of artificial selection. Some of the words have been omitted. Write the most appropriate term in each space. When humans purposefully apply selection to members of a population, the process is known as artificial selection. For example, people have tried to ‘improve’ their cattle for thousands of years. It is desirable for a dairy farmer to have cows with a high milk yield. The farmer will select cows with high milk yields and mate them with bulls whose have high milk yields. Some of the conferring high milk yield are passed onto their female offspring who are then chosen for breeding. This will continue for many . Artificial selection can have disadvantages such as depression which can lead to infertility.
9700_w11_qp_43
THEORY
2011
Paper 4, Variant 3
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Spraying insecticide on the walls inside houses is the main method of controlling a species of Anopheles mosquito in rural India. A number of different insecticides have been used. Malathion was the main insecticide used for many years. In 2005 the newer insecticide, deltamethrin, was used instead and the use of malathion was discontinued. A laboratory study was carried out using mosquitoes collected from two sites in India. The percentage of mosquitoes killed by malathion and deltamethrin was estimated. The results of the study are shown in Table 4.1 Table 4.1 site year percentage of mosquitoes killed by malathion deltamethrin Jamnagar Bikaner With reference to Table 4.1, describe the difference in effectiveness of the two insecticides. The researchers concluded that at Jamnagar, the mosquitoes had evolved resistance to deltamethrin. Explain how the mosquitoes evolved resistance. Explain how the data in Table 4.1 show evidence that the use of malathion was discontinued after 2005. The resistance of mosquitoes to malathion was found to be due to a difference in the shape of one enzyme. Name the type of variation controlling malathion resistance in the mosquito population. Some students suggested that resistance to malathion could be due to a gene with two alleles. They proposed that the allele for resistance to malathion would be dominant to the allele for non-resistance. Using this assumption, the data in Table 4.1 can be used to calculate the frequency of resistant mosquitoes and the frequency of the allele for resistance in a mosquito population. Use the Hardy-Weinberg principle to calculate the frequency, p, of the allele for resistance in Jamnagar in 2005.
9700_w16_qp_41
THEORY
2016
Paper 4, Variant 1
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Spraying insecticide on the walls inside houses is the main method of controlling a species of Anopheles mosquito in rural India. A number of different insecticides have been used. Malathion was the main insecticide used for many years. In 2005 the newer insecticide, deltamethrin, was used instead and the use of malathion was discontinued. A laboratory study was carried out using mosquitoes collected from two sites in India. The percentage of mosquitoes killed by malathion and deltamethrin was estimated. The results of the study are shown in Table 4.1 Table 4.1 site year percentage of mosquitoes killed by malathion deltamethrin Jamnagar Bikaner With reference to Table 4.1, describe the difference in effectiveness of the two insecticides. The researchers concluded that at Jamnagar, the mosquitoes had evolved resistance to deltamethrin. Explain how the mosquitoes evolved resistance. Explain how the data in Table 4.1 show evidence that the use of malathion was discontinued after 2005. The resistance of mosquitoes to malathion was found to be due to a difference in the shape of one enzyme. Name the type of variation controlling malathion resistance in the mosquito population. Some students suggested that resistance to malathion could be due to a gene with two alleles. They proposed that the allele for resistance to malathion would be dominant to the allele for non-resistance. Using this assumption, the data in Table 4.1 can be used to calculate the frequency of resistant mosquitoes and the frequency of the allele for resistance in a mosquito population. Use the Hardy-Weinberg principle to calculate the frequency, p, of the allele for resistance in Jamnagar in 2005.
9700_w16_qp_42
THEORY
2016
Paper 4, Variant 2
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Meiosis is one process that contributes to genetic variation. State precisely the stage of meiosis where single chromosomes line up on the equator. Outline the events taking place during anaphase I of meiosis. Describe how crossing over during meiosis leads to genetic variation. Mutation also causes genetic variation. Some populations of water hemp, Amaranthus tuberculatus, have evolved herbicide resistance as a result of a mutation. This is a problem for farmers as water hemp grows in crop fields, lowering productivity. Two populations of water hemp were tested for resistance to the herbicide mesotrione. One was a population known to be resistant and the other was a test population, whose resistance was unknown. • Leaves were removed and immersed in a radioactively labelled solution of mesotrione. • The leaves absorbed some mesotrione and became radioactive. • Resistant leaves are able to degrade mesotrione by metabolism. • The time for 50% of absorbed mesotrione to degrade was calculated by measuring the radioactivity of the leaves. The results are shown in Table 3.1. Table 3.1 population of water hemp mean time for 50% of absorbed mesotrione to degrade / hours standard deviation test 27.5 4.75 control 10.1 2.34 Explain how the results in Table 3.1 show that the two populations differ in their resistance to mesotrione. Explain why this example of genetic variation is important for natural selection in water hemp populations. Farmers can send in a sample of leaves of water hemp from their fields to a laboratory to be tested for resistance to mesotrione or other herbicides. Suggest the benefit of this to a farmer. The null hypothesis states there is no significant difference between the mean times for 50% of absorbed mesotrione to degrade in the two populations. A t‑test can be carried out to compare these two means. The critical value for t at the p = 0.05 significance level is 2.23. Use the formula in to calculate the value of t. Show your working. t n s n s x x = + - e o Key x = mean s = standard deviation n1 = 6 (number of readings for test population) n2 = 6 (number of readings for control population) t = Use your calculated value of t to explain whether the null hypothesis should be accepted or rejected. accept or reject explanation
9700_w18_qp_41
THEORY
2018
Paper 4, Variant 1
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Questions Discovered
104