16.2. The roles of genes in determining the phenotype
A subsection of Biology, 9700, through 16. Inheritance
Listing 10 of 116 questions
The tiger barb, Puntigrus tetrazona, is a South American fish that is popular worldwide as an aquarium fish. shows the appearance of a normal (wild-type) tiger barb. • Tiger barbs that show a wild-type phenotype are gold with black stripes. • Tiger barbs that show an albino phenotype are gold with white stripes. • In 2012, a fish breeder discovered a tiger barb with a new, transparent, phenotype. This fish had a transparent body and black stripes. The fish breeder crossed the tiger barb showing the new transparent phenotype with a tiger barb showing the albino phenotype. All the F1 offspring were wild-type. These F1 offspring were crossed with each other. Table 6.1 shows the phenotypes obtained in the F2 generation and the number of fish showing each phenotype. Table 6.1 F2 phenotype number of fish wild-type (gold with black stripes) albino (gold with white stripes) transparent with black stripes transparent with white stripes The fish breeder concluded that the new transparent phenotype had occurred because of a mutation. Explain how the results in Table 6.1 support this conclusion. State the approximate whole-number ratio shown by the results in Table 6.1. Explain what the results in Table 6.1 show about the genes and alleles that determine the wild-type, albino and the two different transparent phenotypes in tiger barbs. The F1 tiger barbs all looked the same but the F2 offspring showed variation. The F2 offspring showed four different phenotypes. Describe the processes that occurred during meiosis in the F1 fish that allowed this variation to occur.
9700_w23_qp_41
THEORY
2023
Paper 4, Variant 1
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The tiger barb, Puntigrus tetrazona, is a South American fish that is popular worldwide as an aquarium fish. shows the appearance of a normal (wild-type) tiger barb. • Tiger barbs that show a wild-type phenotype are gold with black stripes. • Tiger barbs that show an albino phenotype are gold with white stripes. • In 2012, a fish breeder discovered a tiger barb with a new, transparent, phenotype. This fish had a transparent body and black stripes. The fish breeder crossed the tiger barb showing the new transparent phenotype with a tiger barb showing the albino phenotype. All the F1 offspring were wild-type. These F1 offspring were crossed with each other. Table 6.1 shows the phenotypes obtained in the F2 generation and the number of fish showing each phenotype. Table 6.1 F2 phenotype number of fish wild-type (gold with black stripes) albino (gold with white stripes) transparent with black stripes transparent with white stripes The fish breeder concluded that the new transparent phenotype had occurred because of a mutation. Explain how the results in Table 6.1 support this conclusion. State the approximate whole-number ratio shown by the results in Table 6.1. Explain what the results in Table 6.1 show about the genes and alleles that determine the wild-type, albino and the two different transparent phenotypes in tiger barbs. The F1 tiger barbs all looked the same but the F2 offspring showed variation. The F2 offspring showed four different phenotypes. Describe the processes that occurred during meiosis in the F1 fish that allowed this variation to occur.
9700_w23_qp_43
THEORY
2023
Paper 4, Variant 3
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Populations of the moth Biston betularia live in Europe and in North America. The most common phenotype on both continents is a pale wing colour with light‑grey shading (the typical form). A moth phenotype with dark wing colour (the melanic form) also occurs on both continents. shows the typical form of the moth. shows the melanic form of the moth. Two melanic European moths were crossed together. The wing colours of the offspring were 15 typical and 41 melanic. Construct a genetic diagram to explain these results. You may use the symbols A and a to represent the alleles. In a similar experiment, two melanic North American moths were crossed together. The colours of the offspring were 10 typical and 31 melanic. What can be concluded about the allele that causes the melanic form in the moth populations in both continents? ……………… Researchers did not know if the allele causing the melanic form in European moths occurred at the same locus as the allele causing the melanic form in North American moths. To find out, they carried out the following crosses: • Cross 1: European moths that were heterozygous at the European melanic locus only were crossed with North American moths that were heterozygous at the North American melanic locus only. • Cross 2: The melanic and the typical offspring of cross 1 were mated together. Explain why cross 2 is a test cross. … … Complete Table 10.1 to show the predicted results if: • the European and North American melanic alleles are on the same locus (A/a) • the European and North American melanic alleles are on two different loci (A/a and B/b). Table 10.1 same locus (A/a) different loci (A/a and B/b) genotypes of melanic moths from cross 1 proportion of test crosses (cross 2) giving 100% melanic offspring A light trap was used to estimate the total size of a population of B. betularia in a woodland. On night one, 24 moths were captured. These were marked with a small spot of harmless paint. On night two, 29 moths were captured, and 8 of these showed a spot of paint. Use the Lincoln index formula provided to calculate the size of the population. Show your working. N m n n # = Key to symbols: N = estimate of population size n1 = number of individuals captured in first sample n2 = number of individuals (both marked and unmarked) captured in second sample m2 = number of marked individuals recaptured in second sample population size = The boundaries and names shown, the designations used and the presentation of material on any maps contained in this question paper/insert do not imply official endorsement or acceptance by Cambridge Assessment International Education concerning the legal status of any country, territory, or area or any of its authorities, or of the delimitation of its frontiers or boundaries.
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THEORY
2024
Paper 4, Variant 1
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Genetic crosses can be used to investigate patterns of inheritance. A mutation in a gene involved in fruit colour in tomato plants, Solanum lycopersicum, results in the production of yellow fruits instead of red fruits. A genetic cross was carried out between a pure breeding plant with a red fruit and a pure breeding plant with a yellow fruit to produce the F1 generation. All offspring plants produced red fruits. The F1 plants were then crossed with each other and the seeds produced were planted to obtain the F2 generation. Construct a genetic diagram to show the cross of the F1 generation that produced this F2 generation. Use the symbols R and r for the alleles. A theoretical dihybrid cross involves two genes located on different autosomes. Each gene has two alleles, one dominant and one recessive. A parent, homozygous dominant for both genes, is crossed with a parent that is homozygous recessive for both genes. This produces F1 individuals that are then crossed to produce the F2 generation. State the phenotypic ratio of this dihybrid F2 generation and explain why some of these offspring phenotypes are different from the original parental phenotypes. ratio explanation
9700_w24_qp_42
THEORY
2024
Paper 4, Variant 2
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Populations of the moth Biston betularia live in Europe and in North America. The most common phenotype on both continents is a pale wing colour with light‑grey shading (the typical form). A moth phenotype with dark wing colour (the melanic form) also occurs on both continents. shows the typical form of the moth. shows the melanic form of the moth. Two melanic European moths were crossed together. The wing colours of the offspring were 15 typical and 41 melanic. Construct a genetic diagram to explain these results. You may use the symbols A and a to represent the alleles. In a similar experiment, two melanic North American moths were crossed together. The colours of the offspring were 10 typical and 31 melanic. What can be concluded about the allele that causes the melanic form in the moth populations in both continents? ……………… Researchers did not know if the allele causing the melanic form in European moths occurred at the same locus as the allele causing the melanic form in North American moths. To find out, they carried out the following crosses: • Cross 1: European moths that were heterozygous at the European melanic locus only were crossed with North American moths that were heterozygous at the North American melanic locus only. • Cross 2: The melanic and the typical offspring of cross 1 were mated together. Explain why cross 2 is a test cross. … … Complete Table 10.1 to show the predicted results if: • the European and North American melanic alleles are on the same locus (A/a) • the European and North American melanic alleles are on two different loci (A/a and B/b). Table 10.1 same locus (A/a) different loci (A/a and B/b) genotypes of melanic moths from cross 1 proportion of test crosses (cross 2) giving 100% melanic offspring A light trap was used to estimate the total size of a population of B. betularia in a woodland. On night one, 24 moths were captured. These were marked with a small spot of harmless paint. On night two, 29 moths were captured, and 8 of these showed a spot of paint. Use the Lincoln index formula provided to calculate the size of the population. Show your working. N m n n # = Key to symbols: N = estimate of population size n1 = number of individuals captured in first sample n2 = number of individuals (both marked and unmarked) captured in second sample m2 = number of marked individuals recaptured in second sample population size = The boundaries and names shown, the designations used and the presentation of material on any maps contained in this question paper/insert do not imply official endorsement or acceptance by Cambridge Assessment International Education concerning the legal status of any country, territory, or area or any of its authorities, or of the delimitation of its frontiers or boundaries.
9700_w24_qp_43
THEORY
2024
Paper 4, Variant 3
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Array comparative genome hybridisation (aCGH) is a technique involving the use of a microarray to analyse a genome or sections of a genome. Outline the steps required to prepare the genome of an individual so that the genome is ready for analysis using a microarray chip. DiGeorge syndrome is a dominant inherited disease in humans. DiGeorge syndrome is caused by deletion of a large number of nucleotides from chromosome 22. The number of nucleotides deleted varies between individuals in a range from 800 000 to 3 100 000. The largest deletions can cause the removal of up to 46 protein-coding genes from the chromosome. shows the results of aCGH using a microarray specific for the section of chromosome 22 within which the DiGeorge syndrome deletion occurs. The microarray analysed DNA from two individuals: • one with DiGeorge syndrome • one who did not have DiGeorge syndrome (control DNA for comparison). In the aCGH results shown in : • Each small circle represents the results from a single probe on the microarray. • The x-axis shows the position of each probe on chromosome 22. The position is shown as distance along the chromosome in millions of nucleotides. • A result close to 100% fluorescence on the y-axis means that the DNA from the individual with DiGeorge syndrome fluoresces at the same intensity as the control DNA for that probe. • A result close to 50% fluorescence on the y-axis means that the DNA from the individual with DiGeorge syndrome fluoresces half as much as the control DNA for that probe. 16.0 17.0 18.0 position of probe on chromosome 22 / millions of nucleotides 19.0 20.0 21.0 fluorescence of DNA from an individual with DiGeorge syndrome as a percentage of the fluorescence of control DNA With reference to , estimate the number of nucleotides deleted from the affected chromosome 22 in the individual with DiGeorge syndrome. Give your answer to the nearest 100 000 nucleotides. Explain how the microarray technique works to give the results shown in . Suggest why the phenotypes of two individuals with DiGeorge syndrome can be different.
9700_m23_qp_42
THEORY
2023
Paper 4, Variant 2
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Use The summer squash plant produces fruit that are either white or yellow in colour and are either shaped like a disc or a sphere. The dominant phenotypes are white and disc- shaped fruit. Using the symbols A for white and a for yellow and B for disc and b for sphere, draw a genetic diagram to show what proportion of offspring will have yellow and sphere-shaped fruit if a white and disc-shaped fruit plant, heterozygous for both genes, is self-fertilised. Examiner’s Use Sickle cell anaemia is a blood disease that is frequently fatal when homozygous. It is caused by an autosomal recessive allele. Heterozygotes have sickle cell trait and appear normal. Malaria is a potentially fatal infectious disease of the blood caused by the protoctist, Plasmodium. In parts of the world where malaria is endemic the frequency of the sickle cell allele is high. Explain the possible health consequences, in such areas, for a person who is homozygous dominant and for a person who is homozygous recessive for the sickle cell allele. homozygous dominant for the sickle cell allele homozygous recessive for the sickle cell allele. Explain why heterozygotes have a strong selective advantage in areas where malaria occurs.
9700_s06_qp_4
THEORY
2006
Paper 4, Variant 0
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Canavan disease is a non sex-linked inherited condition that causes progressive damage to neurones of the brain. Symptoms of the condition include a loss of motor skills and mental retardation. The symptoms appear in early infancy and many children with this condition die by the age of four years. People with Canavan disease lack an enzyme called aspartoacylase which breaks down N-acetyl aspartate. The build up of N-acetyl aspartate can interfere with the formation of the myelin sheath, particularly in neurones of the brain. Enzymes such as aspartoacylase display specificity. Outline what is meant by specificity of an enzyme. Complete the genetic diagram below to show how an unaffected man and an unaffected woman could produce a child with Canavan disease. key to symbols parental phenotypes unaffected man unaffected woman parental genotypes gametes offspring genotypes offspring phenotypes Explain the importance of the myelin sheath in the functioning of a neurone.
9700_s11_qp_41
THEORY
2011
Paper 4, Variant 1
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Huntington’s Disease (HD) is a severe neurological disorder in which symptoms usually appear after the person has reached sexual maturity. Symptoms include memory loss and changes in personality and mood. HD is caused by a gene mutation on chromosome 4 in which the triplet code CAG is repeated many times. The resulting allele is dominant. Explain what is meant by the terms gene mutation and triplet code. gene mutation triplet code A couple wish to start a family. The man does not have HD but the woman does have the disease. The woman’s father does not have the disease. Complete the genetic diagram below to show the probability of the couple’s first child having HD. key Huntington allele = T normal allele = t parental phenotypes man without HD woman with HD parental genotypes gametes offspring genotypes offspring phenotypes probability of first child having HD
9700_s11_qp_42
THEORY
2011
Paper 4, Variant 2
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Huntington’s Disease (HD) is a severe neurological disorder in which symptoms usually appear after the person has reached sexual maturity. Symptoms include memory loss and changes in personality and mood. HD is caused by a gene mutation on chromosome 4 in which the triplet code CAG is repeated many times. The resulting allele is dominant. Explain what is meant by the terms gene mutation and triplet code. gene mutation triplet code A couple wish to start a family. The man does not have HD but the woman does have the disease. The woman’s father does not have the disease. Complete the genetic diagram below to show the probability of the couple’s first child having HD. key Huntington allele = T normal allele = t parental phenotypes man without HD woman with HD parental genotypes gametes offspring genotypes offspring phenotypes probability of first child having HD
9700_s11_qp_43
THEORY
2011
Paper 4, Variant 3
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Questions Discovered
116