16. Inheritance
A section of Biology, 9700
Listing 10 of 199 questions
Albinism is a condition that results from disruption of a biosynthetic pathway involving the enzyme tyrosinase. It is estimated that 1 in 17 000 people in the world has albinism. A recessive mutation in the TYR gene, which codes for the enzyme tyrosinase, is one cause of albinism. Individuals with this form of albinism are homozygous recessive. Describe the phenotype of a person with albinism. A recessive mutation in a different gene causes a type of albinism that mainly affects the eyes (ocular albinism). A person with this condition has reduced clarity of vision and involuntary eye movements. shows the pattern of inheritance of ocular albinism in one family. The pattern indicates sex-linked inheritance. Key: = unaffected female = carrier female = unaffected male = male with ocular albinism Explain why supports sex-linked inheritance of ocular albinism. Draw a genetic diagram to show how individuals 1 and 2 cannot have a child with ocular albinism. key to symbols parental genotypes gametes offspring genotypes offspring phenotypes Ocular albinism may be caused by a base deletion mutation. This mutation results in a non-functional protein. Explain how a base deletion mutation can result in a non-functional protein. Ocular albinism is a non-progressive disorder and clarity of vision remains stable throughout life. A female has a family history of ocular albinism but she does not have any symptoms. A test to find out if she has the mutant allele is available. Suggest one reason for taking this test and one reason against taking this test. for against
9700_s19_qp_43
THEORY
2019
Paper 4, Variant 3
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Haemoglobin is made of two α‑globin chains and two β‑globin chains. A person may have a mutation in the gene coding for β‑globin. This is due to a base substitution and leads to the production of abnormal β‑globin and therefore abnormal haemoglobin. A person who is homozygous for the mutant allele will have a condition called sickle cell anaemia. Outline the phenotypic effects of having abnormal haemoglobin in a person with sickle cell anaemia. HbA is the allele coding for normal β‑globin. HbS is the allele coding for abnormal β‑globin. The alleles are codominant. Explain what is meant by the alleles being codominant. A person who is heterozygous will have sickle cell trait but may not experience the symptoms of sickle cell anaemia. Suggest why a person with sickle cell trait may not show the symptoms of sickle cell anaemia. Construct a genetic diagram to show the possible offspring for a man and a woman who both have sickle cell trait. parental sickle cell trait × sickle cell trait phenotype parental genotype gametes offspring genotypes offspring phenotypes
9700_s21_qp_43
THEORY
2021
Paper 4, Variant 3
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The jaguar, Panthera onca, is a large cat that lives mainly in South America. The majority of jaguars have light brown fur with black spots, as shown in . Some jaguars have completely black fur, as shown in . The pigments involved in fur colour are produced as a result of biochemical pathways that take place in cells called melanocytes. These pathways are similar to those that occur in human melanocytes. The melanocortin 1 receptor (MC1R) is located on the cell surface membrane of melanocytes and is coded for by the MC1R gene. outlines the processes that occur in jaguar melanocytes. phaeomelanin (light brown) eumelanin MC1R not activated MC1R activated MC1R substance X tyrosine cell surface membrane Name the substance represented by X. When MC1R is activated a second messenger is produced in the cell. Give an example of a second messenger. Substance X is also produced in humans, but a mutation of the TYR gene can result in substance X not being produced. Describe the phenotype of a person with this mutation. The MC1R gene has two alleles and is located on an autosome. • When two jaguars with light brown fur mate all the offspring have light brown fur. • When two jaguars with black fur mate either all the offspring will have black fur or some offspring will have black fur and some will have light brown fur. Using symbols, construct a genetic diagram to show how two jaguars with black fur can produce some offspring with black fur and some offspring with light brown fur.
9700_s22_qp_42
THEORY
2022
Paper 4, Variant 2
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A freshwater fish species, Oryzias latipes, has individuals with four body colour patterns, as shown in Table 5.1. Table 5.1 body colour pattern phenotype red white red with black spots white with black spots Two unlinked genes determine the body colour patterns shown in Table 5.1. One gene controls whether the body colour is red or white: • dominant allele R = red • recessive allele r = white. The other gene controls whether black spots are present or not present: • dominant allele B = with black spots • recessive allele b = without black spots. A fish that is homozygous recessive at both loci is white. Genetic crosses were carried out to investigate the inheritance of the four different body colour patterns. Males that were red with black spots, and homozygous at both loci, were crossed with females that were white. The F1 offspring were all red with black spots. These F1 offspring were then crossed to produce the F2 generation. Table 5.2 shows the observed numbers obtained of each of the four different phenotypes for the F2 generation. Table 5.2 phenotype observed expected O–E (O–E)2 (O –E)2 E red with black spots 281.25 white with black spots 93.75 1.25 1.5625 0.017 red 93.75 2.25 5.0625 0.054 white 31.25 |2 = Table 5.2 compares the observed numbers with the numbers that would be expected in the F2 generation for a normal dihybrid ratio. Calculate |2 for the F2 generation by completing Table 5.2. The formula for |2 is: |2 = Σ (O –E)2 E The critical value at p = 0.05 and 3 degrees of freedom is 7.815. Comment on whether the null hypothesis should be accepted or rejected. Further analysis of the results from the F2 generation in Table 5.2 showed that there were no white males or white males with black spots. In O. latipes, females have two X chromosomes and males have an X and a Y chromosome. It was deduced that, in O. latipes: • the gene that controls body colour is located on the X chromosome and the Y chromosome • the gene that controls whether black spots are present or not is located on an autosome. To produce the F2 generation, red males with black spots, XrYRBb, were crossed with red females with black spots, XRXrBb. Complete the Punnett square in to show the genotypes and phenotypes of the F2 generation. • Use the symbols XR, Xr and YR for the alleles of the gene that controls body colour. • Use the symbols B and b for the alleles of the gene that controls whether black spots are present or not. Some of has been completed for you. female gametes male gametes XRB XrB XRb Xrb XrB XRXrBB XRXrBb XrXrBb female female female red + black spots red + black spots white + black spots YRB XRYRBB XrYRBB XRYRBb XrYRBb male male male male red + black spots red + black spots red + black spots red + black spots Xrb XRXrBb XrXrBb XRXrbb female female female red + black spots white + black spots red + no spots YRb XrYRBb XrYRbb male male red + black spots red + no spots Explain why there are no white males or males that are white with black spots in the F2 generation. In another cross, red males with the genotype XrYRbb were mated with white females with the genotype XrXrbb. All the male offspring were expected to be red and all the female offspring were expected to be white. The observed results showed that the offspring included two red females out of 253 and one white male out of 198. Suggest an explanation for this unexpected result.
9700_s23_qp_43
THEORY
2023
Paper 4, Variant 3
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A maize plant produced a total of 381 grains, 216 purple and smooth, 79 purple and shrunken, 65 yellow and smooth and 21 yellow and shrunken. Using the symbols A for purple and a for yellow and B for smooth and b for shrunken, draw a genetic diagram to explain these results. Explain why yellow shrunken grains breed true. Use [T A chi-squared test was carried out to test the significance of the differences between the observed and expected results. Table 2.1 Complete the missing spaces in the Table 2.1 Table 2.2 Use the calculated values of chi-squared test and the table of probabilities to find the probability of the observed ratio of phenotypes differing significantly from the expected. State what conclusions may be drawn from the probability found in . [Total : 12] grain observed observed expected expected [obs no. – exp no.]2 phenotype number ratio ratio number ÷ expected no. purple and 10.3 381 × 9/16 = 214 4/214 = 0.019 smooth purple and 3.8 381 × 3/16 = 71 64/71 = 0.901 shrunken yellow and 3.1 smooth yellow and 1.0 shrunken total chi square number value degrees of freedom 0.50 0.20 0.10 0.05 0.02 0.01 0.001 2.37 4.64 6.25 7.82 9.84 11.34 16.27 probability greater than
9700_w05_qp_4
THEORY
2005
Paper 4, Variant 0
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Use The fruit fly, Drosophila melanogaster, feeds on sugars found in damaged fruits. A fly with normal features is called a wild type. It has a striped body and its wings are longer than its abdomen. There are mutant variations such as an ebony coloured body or vestigial wings. These three types of fly are shown in . wild type ebony body vestigial wing Wild type features are coded for by dominant alleles, A for wild type body and B for wild type wings. Explain what is meant by the terms allele and dominant. allele dominant Examiner’s Use Two wild type fruit flies were crossed. Each had alleles A and B and carried alleles for ebony body and vestigial wings. Draw a genetic diagram to show the possible offspring of this cross. Examiner’s Use When the two heterozygous fruit flies in were crossed, 384 eggs hatched and developed into adult flies. A chi-squared (χ2) test was carried out to test the significance of the differences between observed and expected results. χ2 =  (O – E)2 E where  = sum of O = observed value E = expected value Complete the missing values in Table 7.1. Table 7.1 phenotypes of Drosophila melanogaster grey body long wing grey body vestigial wing ebony body long wing ebony body vestigial wing observed number (O) expected ratio expected number (E) O – E -9 - 4 (O – E)2 (O – E)2 E 0.38 0.22 1.50 Calculate the value for χ2. χ2 = Examiner’s Use Table 7.2 relates χ2 values to probability values. As four classes of data were counted the number of degrees of freedom was 4 – 1 = 3. Table 7.2 gives values of χ2 where there are three degrees of freedom. Table 7.2 probability greater than 0.50 0.20 0.10 0.05 0.01 0.001 values for χ2 2.37 4.64 6.25 7.82 11.34 16.27 Using your value for χ2, and Table 7.2, explain whether or not the observed results were significantly different from the expected results.
9700_w09_qp_41
THEORY
2009
Paper 4, Variant 1
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Use Pompe disease is a rare neuromuscular disease caused by an autosomal recessive allele. This allele prevents the production of an enzyme called acid alpha-glucosidase (AG), which breaks down glycogen in muscle cells. Glycogen can build up in muscle cells causing damage to the cells. This damage leads to muscle weakness which gets worse with time. Explain how two parents, both of whom produce normal amounts of AG, can produce a child with Pompe disease. One form of treatment is enzyme replacement therapy where AG is given through regular injections. Suggest how AG may be manufactured. Name the hormone that stimulates the breakdown of glycogen in liver cells. State under what conditions glycogen would need to be broken down in liver or muscle cells. The MN blood group system is based on the presence of glycoproteins M and N, on the surface membrane of red blood cells, which act as antigens. State what is meant by the term antigen. Examiner’s Use The type of MN antigen on the surface membrane of red blood cells is controlled by a single gene with two alleles, LM and LN. The phenotypes of the MN blood group system are MM, MN and NN. Complete the genetic diagram to show how the MN blood group is inherited. parental phenotypes MN x MN parental genotypes gametes offspring genotypes offspring phenotypes Allele frequencies for LM and LN vary in different human populations throughout the world. Table 7.1 shows the LM and LN allele frequencies from five populations. Table 7.1 population allele frequency / % LM LN Canadian Inuit Egyptian German Chinese Nigerian Discuss the data shown in Table 7.1.
9700_w09_qp_42
THEORY
2009
Paper 4, Variant 2
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Tuberous Sclerosis Complex (TSC) is a genetic condition caused by a dominant allele of the TSC gene, which leads to abnormal growth of tissue in organs such as the heart, lungs and kidneys. Children with TSC can, with treatment, lead reasonably normal lives. About 33% of people with TSC have at least one parent with the condition. Explain the meaning of the terms dominant and gene. dominant gene A couple wish to start a family. The man does not have TSC but the woman does have TSC. The woman’s father does not have the condition. Complete the genetic diagram below to show the probability of the couple’s first child having TSC. key TSC allele = T normal allele = t parental phenotypes man without TSC woman with TSC parental genotypes gametes offspring genotypes offspring phenotypes probability of first child having TSC Suggest how a person may develop TSC when there is no family history of the condition.
9700_w11_qp_41
THEORY
2011
Paper 4, Variant 1
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Tuberous Sclerosis Complex (TSC) is a genetic condition caused by a dominant allele of the TSC gene, which leads to abnormal growth of tissue in organs such as the heart, lungs and kidneys. Children with TSC can, with treatment, lead reasonably normal lives. About 33% of people with TSC have at least one parent with the condition. Explain the meaning of the terms dominant and gene. dominant gene A couple wish to start a family. The man does not have TSC but the woman does have TSC. The woman’s father does not have the condition. Complete the genetic diagram below to show the probability of the couple’s first child having TSC. key TSC allele = T normal allele = t parental phenotypes man without TSC woman with TSC parental genotypes gametes offspring genotypes offspring phenotypes probability of first child having TSC Suggest how a person may develop TSC when there is no family history of the condition.
9700_w11_qp_42
THEORY
2011
Paper 4, Variant 2
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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_41
THEORY
2012
Paper 4, Variant 1
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Questions Discovered
199