16. Inheritance
A section of Biology, 9700
Listing 10 of 199 questions
Sickle cell anaemia is a non-infectious chronic disease. If not treated, sickle cell anaemia can be painful and life-threatening. Sickle cell anaemia is caused by a base substitution mutation in the gene coding for the β-globin polypeptide of haemoglobin. This leads to a change in the primary structure of the polypeptide, as valine is present instead of glutamine. This results in abnormal sickle-shaped red blood cells, which stick together in blood vessels. Symptoms of sickle cell anaemia include painful attacks when red blood cells block capillaries in tissues and organs. Suggest the consequences to cells when sickle-shaped red blood cells block capillaries in tissues and organs. Sickle cell anaemia is an autosomal recessive inherited disorder: • allele HbA codes for the normal β-globin polypeptide • allele HbS codes for the sickle-cell polypeptide. People who are heterozygous (HbA HbS) have sickle cell trait (SCT). For a child to inherit sickle cell anaemia (HbS HbS), both parents must have SCT. A genetic screening program is available for sickle cell anaemia and SCT: • when a mother is screened and found to have SCT, the father is then screened • if the mother becomes pregnant, the fetus is screened for both sickle cell anaemia and for SCT • the test is done either by amniocentesis or by chorionic villus sampling, both of which carry a small risk of the pregnancy failing. Outline two advantages of genetic screening for sickle cell anaemia and SCT. To test for the presence of HbS, DNA is extracted and the polymerase chain reaction (PCR) is carried out with two specific primers. One mutation to produce HbS is a base substitution in the sixth codon of the β-globin gene. The normal codon GAG changes to GTG. The normal-specific primer detects GAG whereas the mutant-specific primer detects GTG. Explain: • why primers are used in PCR • how the use of two specific primers allows the amplification of the normal, sickle cell anaemia and SCT genotypes. Gel electrophoresis is carried out on the products of the PCRs. includes the results for two individuals, A and B, tested for the sickle cell allele. • Each lane has an 860 base pair band to indicate the test is valid. • Lane 1 is a control lane with a 207bp band for an individual with known normal phenotype. • Lane 2 is a control lane with a 207bp band for an individual with known sickle cell anaemia phenotype. • Lanes 1, 3 and 5 contain DNA from the PCR that used normal-specific primer. • Lanes 2, 4 and 6 contain DNA from the PCR that used mutant-specific primer. Lane 1 wells 860bp 207bp Lane 2 Lane 3 Lane 4 individual A individual B Lane 5 Lane 6 Deduce the genotypes and phenotypes of individuals A and B in . A B The number of cases of sickle cell anaemia is highest in sub-Saharan Africa, the Middle East and India. These areas also have a high incidence of malaria. People with SCT are either unaffected or may have mild symptoms of sickle cell anaemia. One advantage of SCT is an increased resistance to malaria. Explain how natural selection operates to maintain the presence of the sickle cell allele in populations in areas with malaria. Parents who use IVF to produce embryos may decide to have embryos genetically screened by a test known as pre-implantation genetic diagnosis (PGD). Only embryos that do not have sickle cell alleles are transferred to the woman’s uterus. Discuss two ethical reasons why parents using IVF may choose not to have PGD.
9700_w20_qp_42
THEORY
2020
Paper 4, Variant 2
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Tyrosinase is an enzyme found in mammals. It is involved in the synthesis of melanin pigment. Mutations in the tyrosinase gene affect a mammal’s hair colour. Table 5.1 compares DNA sequences for codons 974–985 of: • the normal tyrosinase gene of humans • the normal tyrosinase gene of cats that have pigmented hair (normal cat) • the tyrosinase gene of cats that show an albino phenotype (albino cat). The corresponding amino acid sequences of each tyrosinase are shown in the shaded rows. Table 5.1 human CTC CCC TCT TCA GCT GAT GTG GAA TTT TGC CTA AGT normal cat CTC CCC TCC TCT GCT GAT GTG GAA TTT TGC CTA AGT albino cat CTC CCT CCT CTG CTG ATG TGG AAT TTT GCC TAA GTC human Leu Pro Ser Ser Ala Asp Val Glu Phe Cys Leu Ser normal cat Leu Pro Ser Ser Ala Asp Val Glu Phe Cys Leu Ser albino cat Leu Pro Pro Leu Leu Met Trp Asn Phe Ala STOP – A silent mutation involves a base substitution that does not result in an amino acid change. Use Table 5.1 to identify, with reasons, a silent mutation distinguishing humans from normal cats with pigmented hair. State the changes that resulted in the premature STOP codon in the albino cat DNA sequence. Explain why albino cats, homozygous for the mutation that resulted in the premature STOP codon, do not produce melanin. Bioinformatics was used to compare the whole sequence of the tyrosinase genes of humans and cats. Explain why bioinformatics was used to compare these gene sequences and suggest a conclusion that could be made from the percentage similarity data obtained. Siamese cats have a temperature-sensitive tyrosinase that only functions in the cooler areas of the skin. This means they only produce a small quantity of melanin pigment. Melanin is mainly on their ears, face, paws and tail. shows a Siamese cat. The Siamese allele of tyrosinase is recessive to the normal allele that causes full pigmentation all over the body (T) but is dominant to the albino allele . Draw a genetic diagram to show how a cross between an albino cat and a fully pigmented cat can result in offspring that include kittens with Siamese colouring. parent phenotypes: albino × fully pigmented parent genotypes: gametes: F1 genotypes: F1 phenotypes: ratio:
9700_w20_qp_42
THEORY
2020
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_43
THEORY
2020
Paper 4, Variant 3
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The function of the enzyme iduronate 2-sulfatase is to break down certain complex molecules. Hunter syndrome is a rare, inherited condition in humans, caused by iduronate 2-sulfatase not functioning properly or by iduronate 2-sulfatase not being produced. This will cause the complex molecules to build up, leading to progressive damage to many organs. Explain how a gene mutation can lead to iduronate 2-sulfatase not functioning properly or not being produced. Hunter syndrome is caused by a recessive allele located on the X chromosome. Using appropriate symbols, draw a genetic diagram to show the probability of a normal male and a heterozygous normal female having a child with Hunter syndrome. key parental normal male × normal female phenotype parental genotype gametes offspring genotypes offspring phenotypes probability
9700_w21_qp_41
THEORY
2021
Paper 4, Variant 1
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Genome-wide association studies find links between single nucleotide polymorphisms (SNPs) and phenotypic features such as human diseases. SNPs are points on the DNA that vary in the population because of DNA base substitutions. A genome-wide association study investigates the effect of genetic variation on a disease. A large number of people with the disease and a large number of healthy control individuals provide DNA. Microarray chips are used to identify the genotype of each individual at many SNPs. The Wellcome Trust Case Control Consortium (WTCCC) study was an important genome- wide association study. • The study used a microarray chip that identified each person’s genotype at 500 000 different SNPs. • The study looked for links between SNPs and 7 different diseases. • For each disease, 2000 people with the disease were tested. • Their results were compared with the results of 3000 healthy control individuals. Outline how microarrays are used in the analysis of genomes. Explain why bioinformatics was important to the WTCCC study. summarises results for three diseases in the WTCCC study. The 22 human autosomes and the X chromosome (chromosome 23) are shown. Chromosome locations with SNPs that are associated with a disease at a statistically significant level (greater than 5 arbitrary units) are shown in black. 18 20 X 13 14 chromosome number 18 20 X 13 14 chromosome number 18 20 X 13 14 chromosome number level of association / arbitrary units level of association / arbitrary units level of association / arbitrary units rheumatoid arthritis type 1 diabetes type 2 diabetes Identify the chromosomes that contain SNPs that have a high level of association with both rheumatoid arthritis and Type 1 diabetes. With reference to , compare the genetic basis of the three diseases. Individuals can choose to have their DNA analysed on a microarray chip to predict their risk of developing different diseases. Outline the social and ethical considerations of this type of DNA analysis.
9700_w21_qp_42
THEORY
2021
Paper 4, Variant 2
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The function of the enzyme iduronate 2-sulfatase is to break down certain complex molecules. Hunter syndrome is a rare, inherited condition in humans, caused by iduronate 2-sulfatase not functioning properly or by iduronate 2-sulfatase not being produced. This will cause the complex molecules to build up, leading to progressive damage to many organs. Explain how a gene mutation can lead to iduronate 2-sulfatase not functioning properly or not being produced. Hunter syndrome is caused by a recessive allele located on the X chromosome. Using appropriate symbols, draw a genetic diagram to show the probability of a normal male and a heterozygous normal female having a child with Hunter syndrome. key parental normal male × normal female phenotype parental genotype gametes offspring genotypes offspring phenotypes probability
9700_w21_qp_43
THEORY
2021
Paper 4, Variant 3
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The fruit fly, Drosophila melanogaster, has autosomal genes for body colour and wing shape. Gene B/b is involved in the production of body colour: • B = dominant allele for brown body colour • b = recessive allele for black body colour. Gene D/d is involved in wing shape: • D = dominant allele for straight wing • d = recessive allele for curved wing. A dihybrid test cross was carried out between flies heterozygous for body colour and for wing shape and flies homozygous recessive for body colour and for wing shape. Table 5.1 shows the number of offspring of each phenotype obtained in the test cross. Table 5.1 phenotype observed number expected number brown body colour, straight wings brown body colour, curved wings black body colour, straight wings black body colour, curved wings Use Table 5.1 to calculate the expected number of each phenotype if the two genes are on different autosomes. Write your answers in the table. A chi-squared (χ2) test was carried out to compare the observed results with the results that would be expected from a dihybrid cross involving genes on different autosomes. The value of χ2 = 2097.836. Table 5.2 shows the critical values for the χ2 distribution. Table 5.2 degrees of freedom p value 0.05 0.01 0.001 3.841 6.635 10.828 5.991 9.210 13.816 7.815 11.345 16.266 9.488 13.277 18.467 Explain how the value of χ2 and Table 5.2 can be used to assess the significance of the difference between the observed results and the expected numbers in Table 5.1. Suggest explanations for the observed results in Table 5.1.
9700_w23_qp_42
THEORY
2023
Paper 4, Variant 2
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The scientist Gregor Mendel investigated differences in the length of the stem in the pea plant, Pisum sativum. In 1866, he published the results of his investigation into this trait . shows a diagram of a pea plant. internode Mendel observed that the pea plants he grew either had tall stems or dwarf stems. In his investigation, Mendel carried out crosses using pea plants with these two phenotypes. From the results of these crosses, Mendel demonstrated that tall stems were dominant to dwarf stems in pea plants. It is now known that the stem length trait in pea plants is controlled by one gene that has two alleles: • a dominant allele, Le • a recessive allele, le. Describe a cross that could be carried out and how the results of the cross could be analysed to determine the genotype of a pea plant with a tall stem. The scientists P W Brian and H G Hemming identified that the difference in the length of the stem in pea plants was associated with the presence or absence of gibberellin. They published their findings in 1955. Gibberellin leads to a response in plant cells by binding to specific receptor molecules. State the term used to describe a molecule, such as gibberellin, that binds to specific receptor molecules and leads to a response in cells. Suggest the response of the cells in the internode region of the stem, as labelled in , to the presence of gibberellin and describe how this response affects the trait investigated by Mendel.
9700_m24_qp_42
THEORY
2024
Paper 4, Variant 2
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Explain what is meant by the term gene mutation. Rickets is a childhood disorder involving the softening and weakening of bones. It is usually caused by a lack of vitamin D, calcium ions or phosphate ions. A rare form of rickets that cannot be successfully treated with vitamin D therapy is caused by a mutant allele on the X chromosome. shows a pedigree chart for a family that has a history of this condition. key: male without rickets female without rickets male with rickets female with rickets Using the symbols XR for the mutant allele on the X chromosome Xr for the non-mutant allele on the X chromosome state the genotypes of the following individuals. The gene in which this mutation occurs codes for a protein found in the cells of the proximal convoluted tubule of the kidney. This protein is involved in phosphate ion transport across membranes. Suggest why individuals with this mutant allele show symptoms of rickets.
9700_s10_qp_41
THEORY
2010
Paper 4, Variant 1
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Mitochondrial complex I is a large enzyme complex that forms part of the electron transport chain. The enzyme is composed of many different polypeptides. The genes coding for these polypeptides are located either in mitochondrial DNA (mtDNA) or in nuclear DNA. Mutations in these genes can lead to the production of an enzyme that does not function efficiently. This results in a disease known as mitochondrial complex I deficiency. If severe, this can lead to death in early childhood. Explain why people with mitochondrial complex I deficiency may have muscle weakness and difficulty with nervous coordination of movement. When mitochondrial complex I deficiency is caused by mutation in mtDNA: • a cell in an ovary produces gametes with different proportions of normal mitochondria and mitochondria that contain the mtDNA mutation (mutant mitochondria) • a person has disease symptoms when the proportion of mutant mitochondria in their cells exceeds a certain threshold • the severity of disease symptoms, and the age at which they appear, can vary greatly in the children of one woman. In a family with a history of mitochondrial complex I deficiency that is caused by a mutation in a nuclear gene, the probability of a child inheriting the mutation can be predicted. Suggest why, in families where mitochondrial complex I deficiency is caused by mtDNA mutation, it is not possible to predict the probability of a child inheriting the mutation. Genetic screening can be carried out on people with symptoms of mitochondrial complex I deficiency. Previously 7 mtDNA genes and 37 nuclear genes were sequenced. Some of the people tested did not have mutations in any of these genes. As a result, another gene was sequenced in these people and was found to be mutated. This led to the suggestion that genetic screening should sequence a larger proportion of the genome for people suspected of having this disease. Discuss the ethical reasons for and against sequencing a larger proportion of the genome for people suspected of having mitochondrial complex I deficiency. One mutation linked to mitochondrial complex I deficiency is a base substitution. It causes the amino acid glycine to be replaced by the amino acid valine in a region of α helix in a protein that is important for the formation of mitochondrial complex I. Glycine is a small amino acid with an R group of one hydrogen atom whereas valine has a larger and branched R group. Predict how the change in amino acids would affect the structure of the protein.
9700_s20_qp_41
THEORY
2020
Paper 4, Variant 1
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Questions Discovered
199