16. Inheritance
A section of Biology, 9700
Listing 10 of 199 questions
In the fruit fly, Drosophila melanogaster, eye colour and wing shape are controlled by genes. • E/e are alleles of a gene involved in eye colour. • E results in red eyes, e results in purple eyes. • N/n are alleles of a gene involved in wing shape. • N results in normal wings, n results in vestigial (short, non-functional) wings. These genes are both autosomal. A dihybrid cross was carried out between a fly with red eyes and normal wings and a fly with purple eyes and vestigial wings. Both parents were homozygous for both genes. The offspring from the F1 generation were crossed to obtain the F2 offspring. The results are shown in Table 2.1. Table 2.1 F2 offspring phenotype expected number of individuals expected F2 ratio observed number of individuals observed F2 ratio red eye, normal wing 22.1 red eye, vestigial wing 1.0 purple eye, normal wing 1.3 purple eye, vestigial wing 6.4 Complete the missing expected number of individuals and the expected F2 ratio in Table 2.1. A chi-squared test showed that the results for the F2 generation in Table 2.1 were significantly different from those expected. To investigate this, test crosses were carried out using flies taken from the F1 generation and flies that were homozygous recessive for both genes. The investigator assumed that the genes were unlinked and expected a ratio of 1:1:1:1. Draw a genetic diagram of this test cross. Use the symbols E/e and N/n. The results of the test crosses described in are shown in Table 2.2. Table 2.2 offspring phenotype number of individuals red eye, normal wing red eye, vestigial wing purple eye, normal wing purple eye, vestigial wing Flies with red eye, vestigial wing and flies with purple eye, normal wing phenotypes are described as recombinant. Name the stage of meiosis when these recombinants are produced and state how this occurs. Explain why the results in Table 2.2 are different from the expected ratio of 1:1:1:1.
9700_s18_qp_43
THEORY
2018
Paper 4, Variant 3
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The fruit fly, Drosophila melanogaster, feeds on sugars found in damaged fruits. A fruit fly with normal features is described as wild type. It has red eyes and its wings are longer than its abdomen. There are mutant variations such as purple eyes or short wings. shows a wild type fruit fly and a mutant fruit fly with purple eyes and vestigial wings. wild type mutant red eyes long wing purple eyes vestigial wing • The genes coding for eye colour and wing length are located on the same chromosome. • Allele R for red eyes is dominant to allele r for purple eyes. • Allele N for long wings is dominant to allele n for vestigial wings. Explain what is meant by the terms allele and dominant. allele dominant A wild type fruit fly, heterozygous for both genes, was crossed with a fruit fly that was homozygous recessive for both genes. Table 7.1 is a summary of the cross. Complete Table 7.1. Table 7.1 wild type parent double homozygous recessive parent parental phenotype x parental genotype x offspring offspring phenotype offspring genotype number of offspring Explain why the four offspring phenotypes are not present in a 1:1:1:1 ratio.
9700_s21_qp_42
THEORY
2021
Paper 4, Variant 2
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The fruit fly, Drosophila melanogaster, usually has red eyes. A gene for eye colour has four alleles: red, apricot, honey and white. Define the terms gene and allele. gene allele • The allele for red eyes, CR, is dominant to the other three alleles. • The allele for apricot eyes, CA, is dominant to the allele for honey eyes, CH. • The allele for white eyes, CW, is recessive to the other three alleles. Construct a genetic diagram to show the genotypes, phenotypes and ratio of the offspring from a cross between a fruit fly with red eyes, CRCH, and a fruit fly with apricot eyes, CACW. parents phenotypes red eyes apricot eyes parents genotypes CRCH CACW gametes offspring genotypes offspring phenotypes ratio Describe how you would carry out a test cross and use it to determine the genotype of a red- eyed fruit fly.
9700_s22_qp_41
THEORY
2022
Paper 4, Variant 1
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Epistasis occurs when a gene at one locus can affect the expression of a gene at another locus. Define the terms gene and locus. gene locus Fur colour in mice, Mus musculus, is determined by a number of genes. One example is the result of epistatic interaction between two genes, A and B. • Allele A codes for the production of pigment in the fur. • Allele a does not code for the production of pigment and results in white fur . • Allele B codes for the production of brown fur. • Allele b codes for the production of black fur. Construct a genetic diagram to show the results, including the ratio, of a cross between two mice heterozygous for both genes. parent genotypes AaBb x AaBb parent phenotypes gametes ratio White fur is due to a mutation of the TYR gene. This is called albinism. Explain how a mutation of the TYR gene can result in albinism.
9700_s22_qp_43
THEORY
2022
Paper 4, Variant 3
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In mice there are several alleles of the gene that controls the intensity of pigmentation of the fur. The alleles are listed below in order of dominance with C as the most dominant. C = full colour Cch = chinchilla Ch = himalayan Cp = platinum Ca = albino The gene for eye colour has two alleles. The allele for black eyes, B, is dominant, while the allele for red eyes, b, is recessive. A mouse with full colour and black eyes was crossed with a himalayan mouse with black eyes. One of the offspring was albino with red eyes. Using the symbols above, draw a genetic diagram to show the genotypes and phenotypes of the offspring of this cross.
9700_w08_qp_4
THEORY
2008
Paper 4, Variant 0
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In mice, fur colour is controlled by a gene with multiple alleles. These alleles are listed below in no particular order. black and tan = Cbt yellow = Cy agouti = Ca black = Cb Suggest explanations for the results of the following crosses between mice. Mice with agouti fur crossed with mice with black fur may produce all agouti offspring or some agouti and some black offspring. Crosses between heterozygous parents with the genotype Cy Cb always produce a ratio of two yellow mice to one black mouse. Mice with yellow fur crossed with mice with black fur will produce one of the following outcomes: • some yellow offspring and some agouti offspring • some yellow offspring and some black and tan offspring • some yellow offspring and some black offspring. A test cross is used to determine the genotype of an organism. Describe how you would carry out a test cross to determine the genotype of a black and tan mouse.
9700_w12_qp_43
THEORY
2012
Paper 4, Variant 3
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The fruit fly, Drosophila melanogaster, is widely used in genetic research. It has many phenotypic variants in features such as body colour, wing shape and eye colour. Two variations from the normal-winged, grey-bodied phenotype are: • vestigial (very short) wings, coded for by the recessive allele of the gene N/n • ebony body colour, coded for by the recessive allele of the gene G/g. Using the symbols given, state the possible genotypes of normal-winged, grey-bodied fruit flies. Describe how you would determine the genotype of a normal-winged, grey-bodied fly. One of the genes for eye colour is carried on the X chromosome. This gene has different alleles coding for: • red eyes • orange eyes • white eyes. The allele for red eyes (R) is dominant to the allele for orange eyes and dominant to the allele for white eyes . The allele for orange eyes is dominant to that for white eyes. Using these symbols, draw a genetic diagram to show how a cross between a white-eyed male fruit fly with a red-eyed female fruit fly will produce male and female offspring that are either red-eyed or orange-eyed.
9700_w15_qp_43
THEORY
2015
Paper 4, Variant 3
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Questions Discovered
199