16.2. The roles of genes in determining the phenotype
A subsection of Biology, 9700, through 16. Inheritance
Listing 10 of 116 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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Mammals such as sheep, Ovis aries, and goats, Capra hircus, are important agricultural animals that are sometimes kept together in mixed flocks. Very occasionally, live offspring are born from a mating between a male sheep and a female goat. In sheep 2n = 54 and in goats 2n = 60. Calculate the diploid chromosome number of the hybrid offspring of a sheep and a goat. Outline why the classification of sheep and goats suggests that hybridisation between them should not be likely to occur. Normal (wild-type) goats have a gold and black coat colour pattern, known as bezoar, and are also horned (have horns). Domestic goats may have a white coat and may be hornless (do not have horns). These variations are coded for by two unlinked genes: • white coat colour, coded for by the dominant allele of the gene A/a • hornless, coded by the dominant allele of the gene H/h. A cross between a white hornless goat and a bezoar horned goat produced offspring of four different phenotypes. Draw a genetic diagram to show the genotypes of the two parents, their gametes and the offspring, and the phenotypes of the offspring. Horns on agricultural animals such as goats and cattle can be dangerous to the farmer and to other animals. Horns are often prevented from growing in 5-day-old animals by a stressful procedure called disbudding. Genetic modification can cause a deletion in the allele h coding for horns in cattle embryos, so that the allele no longer codes for a functional protein and the embryos grow into cattle that are hornless. State an ethical advantage of this example of genetic modification. Suggest why genetic modification that causes a deletion in the horned allele, in established breeds of dairy cattle, is preferable to selective breeding for hornless animals.
9700_w18_qp_42
THEORY
2018
Paper 4, Variant 2
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Questions Discovered
116