16. Inheritance
A section of Biology, 9700
Listing 10 of 199 questions
The red poppy, Papaver rhoeas, and several species of daisy of the family Compositae often co-exist as weeds of wheat fields. shows changes in the percentage frequency of red poppies and daisies in an area of wheat fields over a six year period from 1998 to 2003. From 1985, the herbicide metsulfuron-methyl was used to control weeds in this area of wheat fields. This practice continued throughout the six year period. 1998 showed the first occurrence of a red poppy known as biotype X. This red poppy had a specific mutation not present in normal red poppies. year percentage frequency daisies normal red poppy biotype X red poppy Describe how the percentage frequencies of daisies and red poppies changed over the six year period. Metsulfuron-methyl acts by inhibiting an enzyme called acetolactate synthetase, which is needed for the daisies and red poppies to synthesise three amino acids essential for growth. The specific mutation carried by the red poppies of biotype X occurred within the gene coding for this enzyme. The mutation changed amino acid 197 of acetolactate synthetase from proline to leucine. Suggest the effect of this mutation on the structure and activity of the acetolactate synthetase enzyme of biotype X red poppies. Suggest the effect of this mutation on the biotype X red poppies in the presence of metsulfuron-methyl. With reference to , predict and explain the effect of biotype X red poppies on the relative proportions of weeds and wheat in the area of wheat fields in 2003 compared to 1998. . . Suggest how stopping the use of the herbicide metsulfuron-methyl and replacing it with a herbicide that inhibits a different target enzyme in weeds would affect the abundance of red poppies of biotype X.
9700_s17_qp_43
THEORY
2017
Paper 4, Variant 3
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Meiosis is described as reduction division. Explain why meiosis is necessary in the life cycle of a sexually reproducing organism. Plants need mineral ions to grow and develop. For example, plants need phosphates and a deficiency inhibits cell division and root growth. Mutations in individuals of some plant populations allow them to survive in mineral-deficient soils. Name two examples of environmental conditions that affect plant phenotype, other than mineral deficiency. For each example, describe how it affects the phenotype. Explain why mutations are important in selection. A study compared root growth of thale cress, Arabidopsis thaliana, in two different soil types: • full nutrient • low phosphate. Two different populations of thale cress were used: • thale cress with a functional enzyme X • thale cress, with a non-functional enzyme X. 30 seedlings from each population were placed in each type of soil and left to grow for seven days. At the start all seedlings had a root of the same length. After seven days, the length of this root was measured again for each seedling. The mean final root length and standard deviation was calculated for each population of thale cress. The mean final root lengths are shown in . full nutrient low phosphate soil type mean final root length / mm Key population with functional enzyme X population with non-functional enzyme X With reference to , describe the effect of the low phosphate soil type compared to the full nutrient soil type on root growth, for both populations of thale cress. The null hypothesis states there is no significant difference between the mean final root lengths of the two populations of thale cress grown in low phosphate soil type. A t-test can be carried out to compare these two means. The critical value for t at the p = 0.05 significance level is 2.00. Table 3.1 population grown in low phosphate soil type mean final root length / mm standard deviation functional enzyme X 0.5 non-functional enzyme X 0.8 shows the formula for calculating the value of t. t n n x x s s = + - f p mean standard deviation sample size (number of measurements) x s n = = = Use the formula in to calculate the value of t. Show your working. t = Use your calculated value of t to explain whether the null hypothesis should be accepted or rejected. accept or reject explanation
9700_s18_qp_43
THEORY
2018
Paper 4, Variant 3
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The patty pan squash plant, Cucurbita pepo, produces edible fruits that vary in colour. The colour of the fruits is controlled by two genes, A/a and B/b, that occur on different chromosomes. • Allele A produces a white fruit colour. • Allele a does not produce a colour by itself but allows the colours coded by gene B/b to show in the phenotype. • Allele B produces a yellow fruit colour. • Allele b produces a green fruit colour. In a dihybrid cross, an AABB plant was crossed with an aabb plant. All the resulting F1 plants produced white fruits. The F1 plants were then crossed with each other to obtain the F2 generation. Complete with the gametes produced by the F1 parents, the F2 genotypes and the F2 phenotypes. State the ratio of fruit colours in the F2 offspring. F1 parents: AaBb × AaBb ratio of fruit colours in F2 offspring Test crosses were carried out on two white-fruited plants, P and Q, from the F2 generation. Each of these plants had its female flowers pollinated with pollen from a green-fruited plant. For plant P, half of the offspring were white and half were yellow. For plant Q, half of the offspring were white and half were green. Deduce the genotypes of plants P and Q. plant P plant Q Plants P and Q show genetic variation with respect to fruit colour alleles. Identify the process that occurred during meiosis in the F1 parents that produced this variation and the stage of meiosis at which it occurred. process stage of meiosis In 1994, two new varieties of squash were grown in the USA. Both varieties showed resistance to two viral diseases of squash plants. • The variety ‘Tigress’ was developed by selective breeding. • The variety ‘Freedom II’ was developed by adding genes for viral coat proteins to the squash genome. Identify, with reasons, the variety of squash that can be described as recombinant. Outline one social advantage and one social disadvantage of the recombinant squash variety compared to non-resistant squash varieties. advantage disadvantage
9700_s20_qp_41
THEORY
2020
Paper 4, Variant 1
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The summer squash plant, Cucurbita pepo, produces edible fruits that vary in shape. shows the fruits of three different varieties of squash plants. Alfresco Di Nizza Patty pan (disc-shaped) Fruit shape in squashes is controlled by two genes, A/a and B/b, that are located on different chromosomes. • A disc-shaped fruit is produced when both dominant alleles, A and B, are present. • A spherical fruit is produced when either allele A or allele B is present, but not if both A and B are present. • A long fruit is produced when both allele A and allele B are absent. Table 2.1 shows the possible genotypes of the Patty pan and Alfresco varieties. Complete Table 2.1 to show the possible genotypes of the Di Nizza variety. Table 2.1 variety possible genotypes Patty pan (disc-shaped) AABB AaBB AABb AaBb Di Nizza Alfresco aabb A gardener used pollen from a male flower of Alfresco to pollinate a female flower of Di Nizza. The gardener grew the seeds produced from this cross and found that half the offspring produced spherical fruits and half produced long fruits. Draw one genetic diagram to explain this result. parent genotypes gametes offspring genotypes offspring phenotypes The offspring show genetic variation with respect to fruit shape alleles. Name the process that occurred during meiosis in the parents that produced this variation and state the stage of meiosis at which it occurred. process stage of meiosis Genetically modified (GM) summer squash plants with resistance to viral diseases have been grown in the USA since 1995. Scientists have been concerned that viral resistance genes pass easily from GM squash plants to their wild relative, the Texas gourd, Cucurbita texana. Explain why the possibility of gene flow from GM squash plants to the Texas gourd is a cause of social and environmental concern. social environmental A study compared the survival of two different types of hybrids that were formed by cross-pollination between GM virus-resistant squash plants and wild Texas gourd plants: • virus-resistant hybrids that had inherited the viral resistance gene • non-resistant hybrids that had not inherited the viral resistance gene. Viral disease outbreaks caused many of the non-resistant hybrids to die. Leaf-eating beetles then moved in larger numbers to the surviving healthy virus-resistant hybrids. The beetles carried a pathogenic bacterium Erwinia which was capable of killing the plants. compares infection with Erwinia in the virus-resistant hybrids and the non-resistant hybrids. percentage of plants infected with Erwinia July August Key: non-resistant hybrids virus-resistant hybrids Discuss whether these results provide support for the use of genetically modified organisms (GMOs) in food production. Suggest why the GM virus-resistant squashes grown by farmers rarely suffer infection by Erwinia.
9700_s20_qp_42
THEORY
2020
Paper 4, Variant 2
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The courgette plant, Cucurbita pepo, produces edible fruits that vary in colour and shape. Fruit colour in courgettes is controlled by the gene A/a. Fruit shape in courgettes is controlled by the gene B/b. • A yellow fruit is produced when the dominant allele A is present. • A round fruit is produced when the dominant allele B is present. Genes A/a and B/b occur on different chromosomes. Table 1.1 shows the genotypes and phenotypes of four different varieties of courgette with respect to their fruit colour and shape. Table 1.1 name of variety genotype fruit colour fruit shape Defender aabb green long Floridor AABB yellow round Golden Dawn AAbb yellow long Tondo di Piacenza aaBB green round The varieties Golden Dawn and Tondo di Piacenza were grown in the same garden and cross-pollination occurred between them. The gardener grew these cross-pollinated F1 seeds into plants that formed fruits. The gardener did not know the genotypes of the parent plants and did not know that cross-pollination had occurred. State the phenotype of the fruits of the F1 plants and explain why it was unexpected for the gardener. The gardener crossed two of these F1 plants. Complete with the F1 gametes, F2 genotypes and F2 phenotypes. State the ratio of fruit phenotypes in the F2 offspring. ratio of fruit phenotypes Watermelons, Citrullus lanatus, are plants in the same family as courgettes. They produce large round edible fruits that usually contain many hard seeds. Seeds are the structures formed when the male and female gametes fuse at fertilisation. In the 1990s a triploid (3n) watermelon plant was developed. To produce the triploid watermelon plant, a normal diploid parent plant (2n = 22) was crossed with an artificially created tetraploid plant (4n = 44). Triploid watermelon plants develop edible fruits but these are sterile and do not contain seeds, making them more enjoyable to eat. Explain why the fruits of the triploid plants are sterile and do not contain seeds. Watermelons are attacked by watermelon mosaic virus (WM. In 1965, a WMV-resistant plant in the same family, Cucurbita ecuadorensis (2n = 40), was found growing wild in South America. State why a WMV-resistant variety of watermelon cannot be obtained by breeding Cucurbita ecuadorensis with a normal diploid watermelon. Outline how the WMV-resistant trait of Cucurbita ecuadorensis could be transferred to watermelon plants.
9700_s20_qp_43
THEORY
2020
Paper 4, Variant 3
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Use Modern varieties of wheat have developed from numerous hybridisation events between different species of wild grasses. shows some of the possible steps that are believed to have been involved in the development of bread wheat, Triticum aestivum. The letters A, B and C represent three different sets of seven chromosomes. einkorn Triticum urartu 14 chromosomes AA × goat grass 1 Aegilops speltoides 14 chromosomes BB hybridisation and doubling of chromosome number emmer wheat Triticum turgidum 28 chromosomes AABB × goat grass 2 Aegilops tauschii 14 chromosomes CC hybridisation and doubling of chromosome number bread wheat Triticum aestivum 42 chromosomes Complete by writing letters to represent the sets of chromosomes in bread wheat. Write your answer on . Examiner’s Use Explain why hybridisation between emmer wheat and goat grass 2 would have produced a sterile hybrid, if doubling of chromosome number had not occurred. With reference to , suggest why Triticum urartu and Triticum turgidum are classified as different species. Triticum turgidum emerged as a new species without being geographically isolated from Triticum urartu. Outline how geographical isolation may result in speciation.
9700_w09_qp_41
THEORY
2009
Paper 4, Variant 1
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A recent development in fertility treatment is called in-vitro maturation (IVM). This is both cheaper and safer than the standard procedure used in in-vitro fertilisation (IVF), especially for women with polycystic ovaries. Hormone treatment can be dangerous for women with this condition, in which a number of ovarian follicles mature at the same time. IVF and IVM are compared in . In-vitro fertilisation (IVF) Women are given drugs for three weeks to suppress ovulation followed by hormones for two weeks. Mature secondary oocytes are harvested. In-vitro maturation (IVM) Immature secondary oocytes are harvested. Immature secondary oocytes are matured in laboratory culture for 24 to 48 hours. Oocytes from about 50% of women are fertilised by mixing the oocytes with a sample of sperm. Oocytes from about 50% of women are fertilised by injecting a single sperm into the cytoplasm of each oocyte. Fertilised oocytes are placed into the uterus. All oocytes are fertilised by injecting a single sperm into the cytoplasm of each oocyte. With reference to , explain why women are treated with hormones for two weeks after being given drugs to suppress ovulation at the beginning of IVF treatment. State the roles of mitosis and meiosis in producing an immature secondary oocyte. Suggest one advantage and one possible disadvantage of fertilising an oocyte by injecting a sperm into its cytoplasm instead of mixing the oocyte with a sample of sperm. advantage disadvantage
9700_w10_qp_41
THEORY
2010
Paper 4, Variant 1
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A recent development in fertility treatment is called in-vitro maturation (IVM). This is both cheaper and safer than the standard procedure used in in-vitro fertilisation (IVF), especially for women with polycystic ovaries. Hormone treatment can be dangerous for women with this condition, in which a number of ovarian follicles mature at the same time. IVF and IVM are compared in . In-vitro fertilisation (IVF) Women are given drugs for three weeks to suppress ovulation followed by hormones for two weeks. Mature secondary oocytes are harvested. In-vitro maturation (IVM) Immature secondary oocytes are harvested. Immature secondary oocytes are matured in laboratory culture for 24 to 48 hours. Oocytes from about 50% of women are fertilised by mixing the oocytes with a sample of sperm. Oocytes from about 50% of women are fertilised by injecting a single sperm into the cytoplasm of each oocyte. Fertilised oocytes are placed into the uterus. All oocytes are fertilised by injecting a single sperm into the cytoplasm of each oocyte. With reference to , explain why women are treated with hormones for two weeks after being given drugs to suppress ovulation at the beginning of IVF treatment. State the roles of mitosis and meiosis in producing an immature secondary oocyte. Suggest one advantage and one possible disadvantage of fertilising an oocyte by injecting a sperm into its cytoplasm instead of mixing the oocyte with a sample of sperm. advantage disadvantage
9700_w10_qp_42
THEORY
2010
Paper 4, Variant 2
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Questions Discovered
199