17.2. Natural and artificial selection
A subsection of Biology, 9700, through 17. Selection and evolution
Listing 10 of 69 questions
There are more than 600 plant species in the genus Ipomoea. Many species are grown for their attractive flowers, and some species are used as crop plants. shows Ipomoea purpurea, the common morning glory. The gene that determines flower colour in I. purpurea has two alleles: • a dominant allele that results in purple flowers • a recessive allele that results in red flowers. A student recorded the flower colour of all the I. purpurea plants in a field. The student recorded 660 plants with purple flowers and 440 plants with red flowers. Assuming the Hardy-Weinberg principle applies to this population, calculate the number of plants in the field that are heterozygous. Use the equations: p + q = 1 p2 + 2pq + q2 = 1 Show your working and give your answer to the nearest whole number. number of heterozygous plants The Japanese morning glory, I. nil, has over 20 different flower colour phenotypes, including shades of blue, purple, red and pink. The flower colour of I. nil is controlled by at least four genes. The flower colour can change gradually after the flowers open each morning and can change with fluctuations in the carbon dioxide concentration of the surrounding air. A student concluded that the flower colour phenotype in I. nil shows continuous variation. Suggest two reasons why the student made this conclusion. Scientists investigated the response of stomata to changing carbon dioxide (CO2) concentrations in the beach morning glory, I. pes-caprae. The scientists placed I. pes-caprae plants in chambers. They measured the width of open stomata (stomatal apertures) after the plants had been exposed to different CO2 concentrations for 40 minutes. Light intensity and temperature were kept constant. The relationship between CO2 concentration and the mean width of stomatal apertures is shown in . mean width of stomatal aperture / μm CO2 concentration / μmol mol–1 0.0 0.5 1.0 1.5 2.0 In 2016, a study measured the atmospheric CO2 concentration as 400 μmol mol–1. In the future, climate change may reduce water availability and increase atmospheric CO2 concentrations in some habitats. Suggest how the stomatal response shown in would allow I. pes-caprae to survive the effects of climate change. Under certain conditions, the closure of stomata is controlled by abscisic acid. Describe how abscisic acid causes the closure of stomata. Scientists are researching whether abscisic acid can be used in crop treatment to increase yield. Evidence suggests that abscisic acid modifies the effect of auxin on elongation growth in plants. Scientists investigated the effect of different concentrations of abscisic acid on root elongation in seedlings of thale cress, Arabidopsis thaliana. The seedlings were divided into four groups: • a control group (0.0 μmol abscisic acid) • three experimental groups, each treated with a different concentration of abscisic acid: 0.1 μmol, 1.0 μmol, or 10.0 μmol. For each group of seedlings, root length was measured for six days during treatment. The rate of root elongation was calculated each day. The results are shown in . 0.1 μmol of abscisic acid day rate of root elongation / μm h–1 1.0 μmol of abscisic acid 10.0 μmol of abscisic acid Key control (no abscisic acid) With reference to , describe the effect of treatment with abscisic acid on the rate of root elongation. The passage outlines the role of auxin in elongation growth in plants. Complete the passage by using the most appropriate scientific terms. The binding of auxin to receptors causes to be pumped into cell walls. This activates proteins called expansins, which disrupt the links between microfibrils. The cell walls are then able to expand.
9700_w23_qp_41
THEORY
2023
Paper 4, Variant 1
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There are more than 600 plant species in the genus Ipomoea. Many species are grown for their attractive flowers, and some species are used as crop plants. shows Ipomoea purpurea, the common morning glory. The gene that determines flower colour in I. purpurea has two alleles: • a dominant allele that results in purple flowers • a recessive allele that results in red flowers. A student recorded the flower colour of all the I. purpurea plants in a field. The student recorded 660 plants with purple flowers and 440 plants with red flowers. Assuming the Hardy-Weinberg principle applies to this population, calculate the number of plants in the field that are heterozygous. Use the equations: p + q = 1 p2 + 2pq + q2 = 1 Show your working and give your answer to the nearest whole number. number of heterozygous plants The Japanese morning glory, I. nil, has over 20 different flower colour phenotypes, including shades of blue, purple, red and pink. The flower colour of I. nil is controlled by at least four genes. The flower colour can change gradually after the flowers open each morning and can change with fluctuations in the carbon dioxide concentration of the surrounding air. A student concluded that the flower colour phenotype in I. nil shows continuous variation. Suggest two reasons why the student made this conclusion. Scientists investigated the response of stomata to changing carbon dioxide (CO2) concentrations in the beach morning glory, I. pes-caprae. The scientists placed I. pes-caprae plants in chambers. They measured the width of open stomata (stomatal apertures) after the plants had been exposed to different CO2 concentrations for 40 minutes. Light intensity and temperature were kept constant. The relationship between CO2 concentration and the mean width of stomatal apertures is shown in . mean width of stomatal aperture / μm CO2 concentration / μmol mol–1 0.0 0.5 1.0 1.5 2.0 In 2016, a study measured the atmospheric CO2 concentration as 400 μmol mol–1. In the future, climate change may reduce water availability and increase atmospheric CO2 concentrations in some habitats. Suggest how the stomatal response shown in would allow I. pes-caprae to survive the effects of climate change. Under certain conditions, the closure of stomata is controlled by abscisic acid. Describe how abscisic acid causes the closure of stomata. Scientists are researching whether abscisic acid can be used in crop treatment to increase yield. Evidence suggests that abscisic acid modifies the effect of auxin on elongation growth in plants. Scientists investigated the effect of different concentrations of abscisic acid on root elongation in seedlings of thale cress, Arabidopsis thaliana. The seedlings were divided into four groups: • a control group (0.0 μmol abscisic acid) • three experimental groups, each treated with a different concentration of abscisic acid: 0.1 μmol, 1.0 μmol, or 10.0 μmol. For each group of seedlings, root length was measured for six days during treatment. The rate of root elongation was calculated each day. The results are shown in . 0.1 μmol of abscisic acid day rate of root elongation / μm h–1 1.0 μmol of abscisic acid 10.0 μmol of abscisic acid Key control (no abscisic acid) With reference to , describe the effect of treatment with abscisic acid on the rate of root elongation. The passage outlines the role of auxin in elongation growth in plants. Complete the passage by using the most appropriate scientific terms. The binding of auxin to receptors causes to be pumped into cell walls. This activates proteins called expansins, which disrupt the links between microfibrils. The cell walls are then able to expand.
9700_w23_qp_43
THEORY
2023
Paper 4, Variant 3
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Maize is an important food crop that has been improved both by selective breeding and by genetic modification. Outline how selective breeding has been used to improve maize. shows part of a maize cob. The cob is made up of many individual seeds called kernels. Each kernel results from a separate fertilisation of a male and a female gamete. Some kernels are yellow and some are purple. yellow kernel purple kernel Name the type of variation shown in . Suggest a genetic explanation for this pattern of variation in colour. type of variation explanation Maize and other crops have been genetically modified since 1996 to produce the Bt toxin to kill insect pests. shows the area of Bt crops grown (plotted points) and the number of insect pest species in which resistance to Bt has been reported . year area of Bt crops grown / million hectares Bt crops resistant species number of resistant pest species Describe and suggest an explanation for the relationship between the area of Bt crops grown and the number of resistant pest species. Suggest one social advantage and one environmental advantage of growing this Bt maize. social advantage environmental advantage
9700_s16_qp_41
THEORY
2016
Paper 4, Variant 1
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Maize is an important food crop that has been improved both by selective breeding and by genetic modification. Outline how selective breeding has been used to improve maize. shows part of a maize cob. The cob is made up of many individual seeds called kernels. Each kernel results from a separate fertilisation of a male and a female gamete. Some kernels are yellow and some are purple. yellow kernel purple kernel Name the type of variation shown in . Suggest a genetic explanation for this pattern of variation in colour. type of variation explanation Maize and other crops have been genetically modified since 1996 to produce the Bt toxin to kill insect pests. shows the area of Bt crops grown (plotted points) and the number of insect pest species in which resistance to Bt has been reported . year area of Bt crops grown / million hectares Bt crops resistant species number of resistant pest species Describe and suggest an explanation for the relationship between the area of Bt crops grown and the number of resistant pest species. Suggest one social advantage and one environmental advantage of growing this Bt maize. social advantage environmental advantage
9700_s16_qp_43
THEORY
2016
Paper 4, Variant 3
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Holstein Friesian cattle are a breed of cattle used by dairy farmers in many countries of the world for the high milk yield of their cows. shows Holstein Friesian cattle. Milk yield in Holstein Friesian cattle is affected by heat stress. Heat stress occurs when homeostatic mechanisms are not enough to keep the body temperature down to normal levels. One of the factors that contributes to heat stress is air temperature. shows: • the mean daily air temperature in Central Europe • the mean monthly milk yield per cow of Holstein Friesian cattle in Central Europe. mean daily air temperature / °C mean monthly milk yield per cow / kg cow–1 month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec key mean daily air temperature standard error (SE) bar mean monthly milk yield per cow With reference to , describe the trends in air temperature and milk yield from April to August. Many dairy farmers in tropical regions use cattle breeds that are tolerant to heat stress (heat-tolerant cattle). These heat-tolerant cattle: • can tolerate higher air temperatures than Holstein Friesian cattle before heat stress occurs • have milder symptoms of heat stress than Holstein Friesian cattle for the same high air temperatures. Where heat stress does not occur, heat-tolerant cattle produce a lower milk yield than Holstein Friesian cattle under the same conditions. Scientists compared DNA sequences of Holstein Friesian cattle and heat-tolerant cattle for a number of genes known to have an effect on body temperature. Twenty genes were found that had alleles associated only with heat-tolerant cattle. With reference to the information provided, including the data in : • state the type of phenotypic variation shown by milk yield in cattle • identify factors that cause phenotypic variation in milk yield in cattle. In each case, give a reason for your choice. type of phenotypic variation and reason for choice factors that cause phenotypic variation and reason for each choice The scientists found that one of the genes studied, PRLR, has a dominant allele known as SLICK. The SLICK allele was identified in Senepol cattle, a heat-tolerant breed, and is not found in Holstein Friesian cattle. Cattle with the SLICK allele have short hair due to reduced hair growth. Scientists have used selective breeding to introduce the SLICK allele into Holstein Friesian cattle. The milk yields of normal Holstein Friesian cattle and Holstein Friesian cattle with the SLICK allele are shown in , during: • March, when the mean daily air temperature is 5 °C. • September, when the mean daily air temperature is 14 °C. mean monthly milk yield per cow / kg cow–1 month March September key Holstein Friesian cattle Holstein Friesian cattle with SLICK allele With reference to , describe the effect of the SLICK allele on milk yield in Holstein Friesian cattle. The SLICK allele differs from the recessive allele by a single nucleotide deletion. This results in a frameshift mutation and introduces a premature stop codon in the PRLR gene. Scientists can use gene editing to replicate this mutation in Holstein Friesian cattle. This provides a way to introduce the SLICK allele into Holstein Friesian cattle without selective breeding. Compare gene editing and selective breeding for introducing the SLICK allele into Holstein Friesian cattle. Include similarities and differences in your answer.
9700_m24_qp_42
THEORY
2024
Paper 4, Variant 2
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shows the male and female flowers of maize. With reference to , describe how the flowering habit of maize encourages wind-pollination. In a maize plant, the anthers normally ripen before the stigmas are mature and ready to receive pollen. This encourages cross-pollination. Explain the potential advantages of cross-pollination to a plant species. Cultivated maize is related to a wild Mexican grass, teosinte. Teosinte looks very different from maize. In particular, teosinte has a very hard layer surrounding the fruit, making it impossible to use as an edible grain. Archaeological excavations have found that maize in its edible form dates back at least 4000 years. It has been argued that the structural differences between teosinte and maize grains are so great that it is unlikely that maize could have been bred from teosinte so long ago. Investigations have been carried out into the genetic differences between teosinte and maize. A gene on chromosome 4, tga 1, containing 1042 base pairs, was found to be responsible for all of the structural differences between teosinte and maize. shows the DNA base sequence of the only part of tga 1 that always differs between teosinte and maize. teosinte GAT TGG GAT CTC AAG GCG GCG GGC GCG TGG maize GAT TGG GAT CTC AAC GCG GCG GGC GCG TGG Outline the principles of electrophoresis as used in sequencing this DNA. Suggest how the difference in the base sequence of the tga 1 gene shown in could cause large differences in phenotype between teosinte and maize. With reference to , explain how these results support the suggestion that it would have been relatively easy for early farmers in Mexico to have bred maize from teosinte.
9700_s08_qp_4
THEORY
2008
Paper 4, Variant 0
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Domestic goats are small, herbivorous animals that provide milk for human use. This is an important source of food for people in rural South Africa. Three Northern European goat varieties have been imported to South Africa because they have higher milk yields than the native South African goats. Table 2.1 compares the mean daily milk yields of these three breeds of Northern European goat in three locations, Northern Europe, Barbados and South Africa. Table 2.1 goat breed mean daily milk yield / kg in different locations Northern Europe Barbados South Africa British Alpine 4.09 2.55 0.75 Saanen 5.17 1.73 1.45 Toggenburg 4.54 3.46 0.56 Native South African goats have a mean daily milk yield of 0.25 kg. Calculate how many times greater the mean daily milk yield will be if a native South African goat is replaced by the Northern European goat breed that gives maximum yield. Show your working and write your answer to two significant figures. answer Explain how the data in Table 2.1 support the claim that some of the variation in mean daily milk yield in goats is due to genetic causes. The climate, vegetation and availability of veterinary care for goats in Northern Europe, Barbados and South Africa are different. Explain how Table 2.1 shows that environmental factors can cause variation in mean daily milk yield in goats. Native South African goats are better adapted to the local conditions in South Africa than a Northern European breed, such as the Saanen. However Saanen goats have the potential for very high milk yield. Outline a programme of selective breeding that could produce a goat with a high milk yield that is adapted to the local conditions in South Africa. Children in developing countries may drink unpasteurised goats’ milk. Some may develop diarrhoea caused by live bacteria ingested in the milk. Scientists have used genetic engineering to develop goats that produce human lysozyme in their milk. Lysozyme is an enzyme that kills bacteria and so reduces the number of bacteria in the milk. State a social advantage and a social disadvantage of making these GM goats available in developing countries. advantage disadvantage
9700_w20_qp_42
THEORY
2020
Paper 4, Variant 2
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Selective breeding has been used to improve the characteristics of crop plants, including maize and rice. Outline how vigorous, uniform varieties of maize were developed by selective breeding. IR8 is a variety of rice plant that was developed in the 1960s. IR8 was developed by breeding together two different varieties of rice: • PETA, which produces a high yield • DGWG, which is a dwarf variety. is a diagram showing the three varieties of rice. IR8 PETA DGWG edible grains water line Suggest why IR8 is an improvement on the PETA variety of rice. In 2009, scientists produced a new variety of rice, IR64-Sub1, by breeding together: • FR13A, a variety which has a low yield but has an allele for flood tolerance • IR64, a variety which produces a high yield. The scientists sequenced the DNA of these three rice varieties. Suggest the benefit of sequencing the DNA of IR64-Sub1. After a few generations of breeding, the scientists crossed IR64-Sub1 with IR64. Explain why the scientists crossed IR64-Sub1 with IR64. Thale cress, Arabidopsis thaliana, is another plant that has dwarf varieties. Scientists treated three varieties of A. thaliana with gibberellin: • the wild type (normal, non-dwarf variety) • dwarf variety A • dwarf variety B. shows the responses of the three varieties to treatment with gibberellin. normal growth wild type dwarf variety A dwarf variety B after gibberellin treatment normal growth after gibberellin treatment normal growth after gibberellin treatment With reference to , suggest explanations for the different responses to gibberellin shown by dwarf variety A and dwarf variety B. BZR1 is a transcription factor that helps to regulate growth and development in A. thaliana. Outline the features of a transcription factor such as BZR1.
9700_w22_qp_41
THEORY
2022
Paper 4, Variant 1
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Selective breeding has been used to improve the characteristics of crop plants, including maize and rice. Outline how vigorous, uniform varieties of maize were developed by selective breeding. IR8 is a variety of rice plant that was developed in the 1960s. IR8 was developed by breeding together two different varieties of rice: • PETA, which produces a high yield • DGWG, which is a dwarf variety. is a diagram showing the three varieties of rice. IR8 PETA DGWG edible grains water line Suggest why IR8 is an improvement on the PETA variety of rice. In 2009, scientists produced a new variety of rice, IR64-Sub1, by breeding together: • FR13A, a variety which has a low yield but has an allele for flood tolerance • IR64, a variety which produces a high yield. The scientists sequenced the DNA of these three rice varieties. Suggest the benefit of sequencing the DNA of IR64-Sub1. After a few generations of breeding, the scientists crossed IR64-Sub1 with IR64. Explain why the scientists crossed IR64-Sub1 with IR64. Thale cress, Arabidopsis thaliana, is another plant that has dwarf varieties. Scientists treated three varieties of A. thaliana with gibberellin: • the wild type (normal, non-dwarf variety) • dwarf variety A • dwarf variety B. shows the responses of the three varieties to treatment with gibberellin. normal growth wild type dwarf variety A dwarf variety B after gibberellin treatment normal growth after gibberellin treatment normal growth after gibberellin treatment With reference to , suggest explanations for the different responses to gibberellin shown by dwarf variety A and dwarf variety B. BZR1 is a transcription factor that helps to regulate growth and development in A. thaliana. Outline the features of a transcription factor such as BZR1.
9700_w22_qp_43
THEORY
2022
Paper 4, Variant 3
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Questions Discovered
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