16. Inheritance
A section of Biology, 9700
Listing 10 of 199 questions
A group of plants, known as Rapid Cycling Brassicas (RCBs), has been developed for use in schools and colleges for genetics experiments. When RCB seedlings develop they can have either purple stems or non-purple stems. Their seed leaves can be either green or yellow-green. Purple stems and green seed leaves are controlled by dominant alleles. The genes for stem colour and seed-leaf colour are located on separate chromosomes. Explain what is meant by a dominant allele. allele dominant Draw a genetic diagram to show the likely outcome of a cross between two RCB plants which are heterozygous for both stem colour and seed-leaf colour. Use the symbols A / a for stem colour and B / b for seed leaf colour.
9700_w12_qp_42
THEORY
2012
Paper 4, Variant 2
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A mutation in a gene in the fruit fly, Drosophila melanogaster, gives rise to white-eyed flies instead of the normal red-eyed flies. The allele for red eyes (R) is dominant to the allele for white eyes . A student crossed a red-eyed fly with a white-eyed fly. The results are shown in Table 1.1. Table 1.1 phenotype of fly number of offspring red-eyed female red-eyed male white-eyed female white-eyed male In Drosophila, males possess two different sex chromosomes, X and Y, as in humans. Complete the genetic diagram below to show how the results in Table 1.1 could have been produced. red-eyed fly parental phenotypes parental genotypes gametes offspring genotypes offspring phenotypes white-eyed fly red-eyed female white-eyed male The chi-squared (χ2) test can be used to analyse the results in Table 1.1. The expected ratio of red-eyed females to white-eyed males is 1:1. Complete Table 1.2 and use this to calculate a value for chi-squared (χ2). χ2 = Σ(O–E)2 E v = n–1 key Σ = sum of v = degrees of freedom n = number of classes O = observed value E = expected value Table 1.2 phenotype of fly O E O–E (O–E)2 (O–E)2 E red-eyed female white-eyed male χ2 = Use your calculated value of χ2 and the table of probabilities below, to test the significance of the difference between observed and expected results. degrees of freedom probability 0.90 0.50 0.10 0.05 0.02 0.45 2.71 3.84 0.21 1.39 4.61 5.99
9700_w13_qp_43
THEORY
2013
Paper 4, Variant 3
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Phenylketonuria (PKU) is a genetic disease which results in a raised concentration of the amino acid phenylalanine in the blood. If left untreated in a newborn baby, it can lead to brain damage. For this reason, babies may be tested for PKU soon after birth. Usually, excess phenylalanine is converted to the amino acid tyrosine by the enzyme phenylalanine hydroxylase (PAH). PKU can be the result of a recessive mutation of the gene coding for PAH. Explain what is meant by a recessive mutation. Using appropriate symbols, complete the diagram below to show how two parents who do not have PKU can have children with, or without, PKU. key to symbols parental phenotypes without PKU without PKU parental genotypes gametes offspring genotypes offspring phenotypes PKU can be caused when a short length of the RNA produced during transcription of the gene coding for PAH is lost. Suggest what effect this would have on the protein that is subsequently produced.
9700_w14_qp_41
THEORY
2014
Paper 4, Variant 1
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Phenylketonuria (PKU) is a genetic disease which results in a raised concentration of the amino acid phenylalanine in the blood. If left untreated in a newborn baby, it can lead to brain damage. For this reason, babies may be tested for PKU soon after birth. Usually, excess phenylalanine is converted to the amino acid tyrosine by the enzyme phenylalanine hydroxylase (PAH). PKU can be the result of a recessive mutation of the gene coding for PAH. Explain what is meant by a recessive mutation. Using appropriate symbols, complete the diagram below to show how two parents who do not have PKU can have children with, or without, PKU. key to symbols parental phenotypes without PKU without PKU parental genotypes gametes offspring genotypes offspring phenotypes PKU can be caused when a short length of the RNA produced during transcription of the gene coding for PAH is lost. Suggest what effect this would have on the protein that is subsequently produced.
9700_w14_qp_42
THEORY
2014
Paper 4, Variant 2
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Phenylketonuria (PKU) is a genetic disease which results in a raised concentration of the amino acid phenylalanine in the blood. If left untreated in a newborn baby, it can lead to brain damage. For this reason, babies may be tested for PKU soon after birth. In people without PKU, excess phenylalanine is converted to the amino acid tyrosine by the enzyme phenylalanine hydroxylase (PAH). One of the mutations that can lead to PKU is caused by a base substitution in the gene coding for PAH. This results in the replacement of the amino acid arginine by the amino acid glutamine. Table 1.1 shows the triplet base codes for arginine and glutamine. Table 1.1 amino acid DNA triplet arginine GCA GCG GCT GCC TCT TCC glutamine GTT GTC Using the information in Table 1.1, explain how a single base substitution can replace arginine with glutamine in PAH. Suggest the consequence to the structure of PAH as a result of a single base substitution in its gene. For some genetic diseases, the proportion of individuals that have the disease can vary in different parts of the world. Sometimes this is due to different environmental conditions. The number of cases of PKU is approximately ten times higher in Europe than in Sub-Saharan Africa. • Individuals who are carriers of the recessive allele for PKU are more resistant to ochratoxin A, produced by a fungus found on contaminated grain products. • Ochratoxin A has been linked to a form of renal cancer. • The fungi that produce ochratoxin A are less common in dry regions such as Sub-Saharan Africa. Suggest and explain the mechanism which has resulted in the different number of cases of PKU in European and Sub-Saharan populations.
9700_w14_qp_43
THEORY
2014
Paper 4, Variant 3
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The blackcap, Sylvia atricapilla, is a small song bird. It is a summer visitor to parts of northern Europe, where it breeds. Many blackcaps spend the winter in southern Europe, particularly in Spain. As a result of many people putting out food for birds in their gardens, some birds can survive the winter in the UK. Scientists measured the genetic variation between blackcaps from two forest sites in Germany, 800 km apart. Both sites included birds that had overwintered in Spain and in the UK. The measurements were made shortly after the birds returned from their winter feeding grounds. Explain how DNA sequencing can be used to measure the genetic variation of birds. The measurements of genetic variation showed that: • birds that overwinter in the same country (Spain or the UK) shared many alleles, even though they were living 800 km apart in Germany in the summer • birds that overwintered in different countries (Spain or the UK) shared fewer alleles, even though they were living in the same forest in Germany in the summer • the genetic differences between the birds that overwinter in Spain suggest that they no longer breed with those that overwinter in the UK.
9700_w15_qp_41
THEORY
2015
Paper 4, Variant 1
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The blackcap, Sylvia atricapilla, is a small song bird. It is a summer visitor to parts of northern Europe, where it breeds. Many blackcaps spend the winter in southern Europe, particularly in Spain. As a result of many people putting out food for birds in their gardens, some birds can survive the winter in the UK. Scientists measured the genetic variation between blackcaps from two forest sites in Germany, 800 km apart. Both sites included birds that had overwintered in Spain and in the UK. The measurements were made shortly after the birds returned from their winter feeding grounds. Explain how DNA sequencing can be used to measure the genetic variation of birds. The measurements of genetic variation showed that: • birds that overwinter in the same country (Spain or the UK) shared many alleles, even though they were living 800 km apart in Germany in the summer • birds that overwintered in different countries (Spain or the UK) shared fewer alleles, even though they were living in the same forest in Germany in the summer • the genetic differences between the birds that overwinter in Spain suggest that they no longer breed with those that overwinter in the UK.
9700_w15_qp_42
THEORY
2015
Paper 4, Variant 2
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Diabetes insipidus (Dis a condition that results in excessive thirst and the excretion of large amounts of dilute urine. A main cause of DI is a deficiency of ADH. ADH normally binds to receptor proteins on cell surface membranes of cells in the collecting ducts of nephrons. Outline the effect of ADH on the collecting ducts. An inherited form of DI may be caused by faulty membrane receptors. These receptors are coded for by a sex-linked allele. Explain the term sex linkage. Use a genetic diagram to show how parents who do not have this condition can have a child with the inherited form of DI. symbols parental genotypes gametes offspring genotypes offspring phenotypes
9700_w16_qp_41
THEORY
2016
Paper 4, Variant 1
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Diabetes insipidus (Dis a condition that results in excessive thirst and the excretion of large amounts of dilute urine. A main cause of DI is a deficiency of ADH. ADH normally binds to receptor proteins on cell surface membranes of cells in the collecting ducts of nephrons. Outline the effect of ADH on the collecting ducts. An inherited form of DI may be caused by faulty membrane receptors. These receptors are coded for by a sex-linked allele. Explain the term sex linkage. Use a genetic diagram to show how parents who do not have this condition can have a child with the inherited form of DI. symbols parental genotypes gametes offspring genotypes offspring phenotypes
9700_w16_qp_42
THEORY
2016
Paper 4, Variant 2
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Domestic goats are small, herbivorous animals that provide milk for human use. Goats’ milk is an important source of food for people living in rural China. Xinong Saanen and Guanzhong are the names of two varieties of goat common in China. In these breeds, there is genetic variation at nucleotide position 5752 of a gene coding for a growth factor. At this position there is either a cytosine (C) or a guanine (G) nucleotide. Some individuals are homozygous for the allele containing C at this position (CC), some are homozygous for the allele containing G at this position (GG) and some are heterozygous (CG). Table 2.1 compares the mean milk yield of the first milk-producing period (first lactation) and the next milk-producing period (second lactation) for Xinong Saanen goats of each genotype. Table 2.1 genotype at position 5752 mean milk yield / kg first lactation second lactation CC CG GG Variation in a phenotypic characteristic such as milk yield is caused by a combination of genetic and environmental factors. Goats also show variation in milk yield between the first lactation and second lactation. Suggest, with reasons, whether the variation in milk yield between the first lactation and second lactation, as shown in Table 2.1, is genetic or environmental. The variation at position 5752 of the gene coding for a growth factor is due to a substitution mutation from G to C. With reference to Table 2.1, describe the importance of the substitution from G to C. In a population of 268 Xinong Saanen goats: • the frequency of the C allele is 0.30 • the frequency of the G allele is 0.70. The Hardy-Weinberg principle can be used to predict the number of goats with CC, CG and GG genotypes in the population, using the equation: p2 + 2pq + q2 = 1 For example, the number of goats with genotype GG can be predicted to be 131. Use the Hardy-Weinberg principle to predict the number of goats with genotypes CC and CG in this population of Xinong Saanen goats. number of goats with genotype CC number of goats with genotype CG Table 2.2 shows the actual number of goats with each genotype in a population of Xinong Saanen goats and in a population of Guanzhong goats. Table 2.2 population total number of goats number of goats of each genotype allele frequency CC CG GG p q Xinong Saanen 0.70 0.30 Guanzhong 0.81 0.19 A close match between your predicted figures in and the actual numbers in Table 2.2 would mean that the Xinong Saanen population is in Hardy-Weinberg equilibrium. State the name of a statistical test that could be used to find out whether or not the Xinong Saanen population is in Hardy-Weinberg equilibrium. The predicted numbers of goats with each genotype in the Guanzhong population according to the Hardy-Weinberg principle are: • CC = 16 • CG = 135 • GG = 289. These figures are significantly different from the actual figures in Table 2.2. With reference to Table 2.2, describe the evidence that shows that the Guanzhong population is not in Hardy-Weinberg equilibrium and suggest reasons for this. Goats can be genetically modified to produce human proteins in their milk. In 2009, an anti-clotting protein produced in this way was approved for use as a drug in people who lack the protein. State one ethical advantage and one ethical problem of producing medicinal drugs from the milk of genetically modified goats. advantage problem
9700_w20_qp_41
THEORY
2020
Paper 4, Variant 1
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Questions Discovered
199