16. Inheritance
A section of Biology, 9700
Listing 10 of 199 questions
In some varieties of domestic cats the gene for fur colour is located on the X chromosome. This gene has two alleles. One allele codes for black fur and the other allele codes for ginger fur. The two alleles are codominant so a heterozygous cat will have fur with patches of black and ginger colours. A cat with fur of two colours is known as a tortoiseshell. Using appropriate symbols, construct a genetic diagram to show the results of a cross between a female tortoiseshell cat and a male ginger cat. symbols parent phenotypes tortoiseshell female ginger male parent genotypes gametes offspring genotypes offspring phenotypes In humans the TYR gene is involved in the production of a dark pigment, melanin, in some cells. Describe how the expression of the TYR gene leads to the production of melanin.
9700_w22_qp_43
THEORY
2022
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.
9700_w24_qp_41
THEORY
2024
Paper 4, Variant 1
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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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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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Resistance to the poison warfarin is now extremely common in rats. Warfarin inhibits an enzyme in the liver, vitamin K epoxide reductase, that is necessary for the recycling of vitamin K. This vitamin is involved in the production of substances required for blood clotting. • Rats susceptible to warfarin die of internal bleeding. • Rats that are homozygous for resistance to warfarin do not suffer from internal bleeding when their diet provides more than 70 μg of vitamin K per kg body mass per day. • Heterozygous rats are resistant to warfarin when their diet provides about 10 μg of vitamin K per kg body mass per day. Using appropriate symbols, complete the genetic diagram to show how two resistant rats can produce warfarin-susceptible offspring. key to symbols parental phenotypes resistant male parental genotypes gametes offspring genotypes offspring phenotypes resistant female Rats that are homozygous for warfarin resistance have a low survival rate in the wild. Suggest why this is so. Warfarin can be safely given to humans who are at risk of unwanted blood clots. The clotting time of the blood is measured regularly and the warfarin dose is varied accordingly. Suggest, giving a reason, the type of inhibition warfarin has on the enzyme vitamin K epoxide reductase. type of inhibition reason The allele for warfarin resistance may have originated by a single base substitution and resulted in a modified vitamin K epoxide reductase. Explain how a single base substitution may affect the phenotype of an organism.
9700_s13_qp_42
THEORY
2013
Paper 4, Variant 2
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Deer mice, Peromyscus maniculatus, are small rodents that live in North America. Like all mammals, their blood contains haemoglobin which combines with oxygen in the lungs, and unloads its oxygen in respiring tissues. Deer mice show variation in their genotypes for the genes that code for the α-polypeptide chain of haemoglobin. In most populations of deer mice, the majority of individuals have the genotype A1A1, while a smaller number have the genotype A0A0. In mice with the genotype A1A1, the amino acid at position 64 in the α-polypeptide chain is aspartic acid. In mice with the genotype A0A0, the amino acid at this position is glycine. Suggest how the change from aspartic acid to glycine in the α-polypeptide chain could have been brought about. The genotypes of deer mice from three different populations, each living at a different altitude, were analysed. shows the relative proportions of deer mice with aspartic acid (white areas) and glycine (black areas) at position 64 in the α-polypeptide of their haemoglobin. altitude 4347 m 1158 m 620 m Describe the effect of altitude on the frequency of the haemoglobin alleles in these populations of deer mice. The partial pressure of oxygen is relatively low at high altitudes. Haemoglobin containing glycine at position 64 in the α-polypeptide chain has a higher affinity for oxygen than haemoglobin with aspartic acid at this position. Suggest how natural selection could account for the difference in allele frequency in deer mice living at high altitudes and low altitudes.
9700_s14_qp_41
THEORY
2014
Paper 4, Variant 1
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Questions Discovered
199