17. Selection and evolution
A section of Biology, 9700
Listing 10 of 104 questions
Gold ions (Au3+) are toxic to most microorganisms. However, the bacterium Delftia acidovorans is frequently found in sticky layers, called biofilms, that form on the surface of gold deposits. D. acidovorans produces a peptide synthase that catalyses the synthesis of a small peptide called delftibactin. When isolated, delftibactin can precipitate Au3+ ions as small particles of metallic gold. Delftibactin is a secondary metabolite. Name another example of a secondary metabolite and explain what is meant by the term. example explanation A mutant strain of D. acidovorans has been identified in which the gene coding for peptide synthase is inactive. The wild-type and mutant D. acidovorans were grown on agar plates and then flooded with gold chloride solution, which contains Au3+ ions. The appearance of such a plate after this treatment is shown in . mutant 'DFLGRYRUDQV wild-type 'DFLGRYRUDQV halo of small particles of gold With reference to , suggest how delftibactin protects D. acidovorans from toxic Au3+ ions. Wild-type and mutant D. acidovorans were grown in standardised conditions: • wild-type and mutant bacteria were grown in the absence of Au3+ ions • wild type and mutant bacteria were grown in the presence of Au3+ ions • mutant bacteria were grown in the presence of Au3+ ions and of delftibactin. The results are shown in . number of bacteria (log scale) in the absence of Au3+ ions wild-type bacteria mutant bacteria in the presence of Au3+ ions in the presence of Au3+ ions and delftibactin Key Explain whether or not the results shown in support the idea that delftibactin is protective. The secondary metabolite, delftibactin, could be used to remove the toxic Au3+ ions that are present in the waste produced by gold mining. Describe how delftibactin could be produced on a large scale.
9700_w15_qp_41
THEORY
2015
Paper 4, Variant 1
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Gold ions (Au3+) are toxic to most microorganisms. However, the bacterium Delftia acidovorans is frequently found in sticky layers, called biofilms, that form on the surface of gold deposits. D. acidovorans produces a peptide synthase that catalyses the synthesis of a small peptide called delftibactin. When isolated, delftibactin can precipitate Au3+ ions as small particles of metallic gold. Delftibactin is a secondary metabolite. Name another example of a secondary metabolite and explain what is meant by the term. example explanation A mutant strain of D. acidovorans has been identified in which the gene coding for peptide synthase is inactive. The wild-type and mutant D. acidovorans were grown on agar plates and then flooded with gold chloride solution, which contains Au3+ ions. The appearance of such a plate after this treatment is shown in . mutant 'DFLGRYRUDQV wild-type 'DFLGRYRUDQV halo of small particles of gold With reference to , suggest how delftibactin protects D. acidovorans from toxic Au3+ ions. Wild-type and mutant D. acidovorans were grown in standardised conditions: • wild-type and mutant bacteria were grown in the absence of Au3+ ions • wild type and mutant bacteria were grown in the presence of Au3+ ions • mutant bacteria were grown in the presence of Au3+ ions and of delftibactin. The results are shown in . number of bacteria (log scale) in the absence of Au3+ ions wild-type bacteria mutant bacteria in the presence of Au3+ ions in the presence of Au3+ ions and delftibactin Key Explain whether or not the results shown in support the idea that delftibactin is protective. The secondary metabolite, delftibactin, could be used to remove the toxic Au3+ ions that are present in the waste produced by gold mining. Describe how delftibactin could be produced on a large scale.
9700_w15_qp_42
THEORY
2015
Paper 4, Variant 2
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The rainbow trout, Oncorhynchus mykiss, is a fish that is bred in commercial fish farms. Rainbow trout that have a blue-silver colour are sold at a higher price than rainbow trout that have a brown colour. The number of fish with each of the different colours was recorded in a breeding population in one fish farm. • population total = 2936 • number of blue-silver fish = 1437 • number of brown fish = 1499 The colour of rainbow trout is controlled by a single autosomal gene with two alleles, one dominant and one recessive. The blue-silver colour occurs when a fish is homozygous for the recessive allele. The formulae of the Hardy–Weinberg principle state that: p + q = 1 p2 + 2pq + q2 = 1 Use these formulae to calculate the expected number of brown fish that are homozygous, if the Hardy–Weinberg principle applies to this population. Show your working. answer = Some scientists suggested that the Hardy–Weinberg principle did not apply in this situation because the rainbow trout were being kept in a commercial fish farm. State two conditions that must be met by this commercial fish farm population for the Hardy–Weinberg principle to apply. Describe how selective breeding can be used to increase the proportion of rainbow trout with a blue-silver colour in commercial fish farms. The Atlantic salmon, Salmo salar, can be genetically modified (GM). GM Atlantic salmon are bred in commercial fish farms for food production. In 2017, GM Atlantic salmon bred in Canada became the first GM animal to enter the human food chain. Explain how the genetic modification of the Atlantic salmon can be used to increase food production.
9700_m20_qp_42
THEORY
2020
Paper 4, Variant 2
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Cystic fibrosis is an autosomal recessive genetic disease. People with cystic fibrosis have a homozygous recessive genotype. Explain the meaning of the terms homozygous and recessive. homozygous recessive In 2020: • there were 10 800 people with cystic fibrosis in the UK • the UK population was estimated to be 67 100 000 people. A proportion of people in the UK population are heterozygous for the gene that causes cystic fibrosis and do not have symptoms of the disease. Use the Hardy–Weinberg principle to calculate the number of people in the UK population who are expected to be heterozygous for the gene that causes cystic fibrosis. The two equations for the Hardy–Weinberg principle are provided. equation 1 p + q = 1 equation 2 p2 + 2pq + q2 = 1 p = frequency of the dominant allele q = frequency of the recessive allele p2 = frequency of the homozygous dominant genotype 2pq = frequency of the heterozygous genotype q2 = frequency of the homozygous recessive genotype The first stage of the calculation has been completed for you. q2 = 10 800 67 100 000 q = p = 2pq = number of people in the UK expected to be heterozygous for the gene = The Hardy–Weinberg principle provides a useful estimate of the number of people in the UK who are heterozygous for cystic fibrosis. However, the estimate is lower than the actual number. This underestimation occurs because not all the conditions of the Hardy–Weinberg principle apply. In the UK in 2020, the mean life expectancy of: • people with cystic fibrosis was approximately 50 years • all people was approximately 80 years. Explain how this information accounts for the underestimation of the number of people in the UK that are heterozygous for cystic fibrosis. A screening programme for cystic fibrosis was introduced in 2007 for all children born in the UK. Children are tested within seven days of their birth. Children identified from the screening programme as being at high risk of having cystic fibrosis can have a genetic test to confirm whether they have the disease. Table 3.1 shows the median predicted life expectancy for people born in the UK who have cystic fibrosis. Predictions are shown for people born in 2008, 2012, 2016 and 2020. Table 3.1 year of birth median predicted life expectancy / years 38.8 43.5 47.0 50.6 Describe the trend shown in Table 3.1 and outline how early screening for cystic fibrosis may have contributed to this trend. In many countries, a genetic test for cystic fibrosis is available to adults who do not have cystic fibrosis but have a family member who either has cystic fibrosis or is heterozygous for the gene that causes cystic fibrosis. These adults include partners, parents, offspring, brothers and sisters of the family member. The aim is to find out if any of these adults are heterozygous for the gene that causes cystic fibrosis. Discuss the ethical and social considerations of making a genetic test for cystic fibrosis available to these adults.
9700_m24_qp_42
THEORY
2024
Paper 4, Variant 2
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Maize originated in the Americas, and 55% of the world’s maize production is from this part of the world. shows the mean yields of maize in the USA between 1860 and 2010. grain yield / tonnes per hectare year Describe the changes in grain yield between 1860 and 2010. The greatest improvement in maize yields came after growers realised that maize hybrids have a much greater yield than inbred lines. Between 1860 and the 1930s, maize was allowed to pollinate naturally in the field. From the 1930s onward, maize seed was produced using ‘double-cross’ hybrids. To produce a double-cross hybrid: • two different maize plants, A and B, are crossed to produce a hybrid, C • two other maize plants, X and Y, are crossed to produce a hybrid, Z • the hybrid C is then crossed with the hybrid Z, to produce the double-cross hybrid. From 1960 onwards, maize seed was produced using ‘single-cross’ hybrids. This involves crossing one inbred (entirely homozygous) plant with a different inbred plant. Explain why single-cross hybrids are genetically uniform, but double-cross hybrids are not. An experiment was carried out in 1996–1997 to investigate the relative effects of genotype and environment on the yield of maize. Maize seeds with different ‘inbreeding coefficients’ were used. The greater the inbreeding coefficient, the greater the degree of homozygosity in the maize plants. Maize seeds with different inbreeding coefficients were planted in two different areas in 1996, and in the same two areas in 1997. shows the results. 0.0 grain yield / tonnes per hectare 0.2 0.4 0.6 inbreeding coefficient 0.8 1996 site 1 1996 site 2 1997 site 1 1997 site 2 1.0 Inbreeding depression is a reduction in vigour that results from inbreeding. Explain how the results in demonstrate inbreeding depression in maize. Explain how the results in show that the environment affects maize yields.
9700_w13_qp_41
THEORY
2013
Paper 4, Variant 1
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Maize originated in the Americas, and 55% of the world’s maize production is from this part of the world. shows the mean yields of maize in the USA between 1860 and 2010. grain yield / tonnes per hectare year Describe the changes in grain yield between 1860 and 2010. The greatest improvement in maize yields came after growers realised that maize hybrids have a much greater yield than inbred lines. Between 1860 and the 1930s, maize was allowed to pollinate naturally in the field. From the 1930s onward, maize seed was produced using ‘double-cross’ hybrids. To produce a double-cross hybrid: • two different maize plants, A and B, are crossed to produce a hybrid, C • two other maize plants, X and Y, are crossed to produce a hybrid, Z • the hybrid C is then crossed with the hybrid Z, to produce the double-cross hybrid. From 1960 onwards, maize seed was produced using ‘single-cross’ hybrids. This involves crossing one inbred (entirely homozygous) plant with a different inbred plant. Explain why single-cross hybrids are genetically uniform, but double-cross hybrids are not. An experiment was carried out in 1996–1997 to investigate the relative effects of genotype and environment on the yield of maize. Maize seeds with different ‘inbreeding coefficients’ were used. The greater the inbreeding coefficient, the greater the degree of homozygosity in the maize plants. Maize seeds with different inbreeding coefficients were planted in two different areas in 1996, and in the same two areas in 1997. shows the results. 0.0 grain yield / tonnes per hectare 0.2 0.4 0.6 inbreeding coefficient 0.8 1996 site 1 1996 site 2 1997 site 1 1997 site 2 1.0 Inbreeding depression is a reduction in vigour that results from inbreeding. Explain how the results in demonstrate inbreeding depression in maize. Explain how the results in show that the environment affects maize yields.
9700_w13_qp_42
THEORY
2013
Paper 4, Variant 2
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In shorthorn cattle, coat colour is controlled by a codominant pair of alleles. Coat colour can be red, white or roan, which is a mixture of red and white. The presence of horns is controlled by a separate pair of alleles. The allele coding for horns is recessive to the allele coding for hornless cattle. Choose suitable symbols for the following: allele for red coat colour allele for white coat colour allele for horns allele for no horns Using the symbols you have chosen, write down the genotypes of: white, hornless cattle roan, horned cattle Some fast-growing breeds of cattle have been produced by artificial selection. Outline the ways in which artificial selection differs from natural selection.
9700_w15_qp_41
THEORY
2015
Paper 4, Variant 1
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In shorthorn cattle, coat colour is controlled by a codominant pair of alleles. Coat colour can be red, white or roan, which is a mixture of red and white. The presence of horns is controlled by a separate pair of alleles. The allele coding for horns is recessive to the allele coding for hornless cattle. Choose suitable symbols for the following: allele for red coat colour allele for white coat colour allele for horns allele for no horns Using the symbols you have chosen, write down the genotypes of: white, hornless cattle roan, horned cattle Some fast-growing breeds of cattle have been produced by artificial selection. Outline the ways in which artificial selection differs from natural selection.
9700_w15_qp_42
THEORY
2015
Paper 4, Variant 2
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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_41
THEORY
2023
Paper 4, Variant 1
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Questions Discovered
104