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Part B Maize Genetics Aleurone Color and Endosperm Characteristics In the corn grain, the aleurone layer is a single layer of cells between the pericarp and the endosperm. The color of the aleurone layer is determined by several interacting genes, one of dominant which, R, is responsible for anthocyanin production in the aleurone layer. Homozygous (RR) and heterozygous (Rr) genotypes result in a purple aleurone layer and the homozygous recessive (rr) genotype results in a colorless one. The grains will appear either purple or yellow rather than purple or colorless as you might expect. The yellow is due to the presence of yellow endosperm below the colorless, transparent aleurone layer. The allele Su produces starchy endosperm and the allele su produces sweet endosperm. Su is completely dominant over su. When the kernels dry, SuSu and Susu grains are smooth, while susu grains are wrinkled. Predict what phenotypic ratio of purple-smooth; purple-wrinkled; yellow-smooth; yellow-wrinkled kernels you expect to observe. 37. What is your expected dihybrid F2 phenotypic ratio 38. What is your null hypothesis? 39. What is your alternative hypothesis?

Part B Maize Genetics Aleurone Color and Endosperm Characteristics In the corn grain, the aleurone layer is a single layer of cells between the pericarp and the endosperm. The color of the aleurone layer is determined by several interacting genes, one of dominant which, R, is responsible for anthocyanin production in the aleurone layer. Homozygous (RR) and heterozygous (Rr) genotypes result in a purple aleurone layer and the homozygous recessive (rr) genotype results in a colorless one. The grains will appear either purple or yellow rather than purple or colorless as you might expect. The yellow is due to the presence of yellow endosperm below the colorless, transparent aleurone layer. The allele Su produces starchy endosperm and the allele su produces sweet endosperm. Su is completely dominant over su. When the kernels dry, SuSu and Susu grains are smooth, while susu grains are wrinkled. Predict what phenotypic ratio of purple-smooth; purple-wrinkled; yellow-smooth; yellow-wrinkled kernels you expect to observe. 37. What is your expected dihybrid F2 phenotypic ratio 38. What is your null hypothesis? 39. What is your alternative hypothesis?

Human Anatomy & Physiology (11th Edition)
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
Section: Chapter Questions
Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...
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34. What are the possible genotypes of the progeny?. What are the phenotypes of the progeny?. c. The F1 plants of problem b above are self-pollinated. (For this problem, assume that no crossing-over has taken place.) 35. What alleles will their gametes carry? eggs 36. What are the genotypes of the F2 population?. What are the phenotypes of the F2 population?_ sperm Part B Maize Genetics Aleurone Color and Endosperm Characteristics In the corn grain, the aleurone layer is a single layer of cells between the pericarp and the endosperm. The color of the aleurone layer is determined by several interacting genes, one of which, R, is responsible for anthocyanin production in the aleurone layer. Homozygous dominant (RR) and heterozygous (Rr) genotypes result in a purple aleurone layer and the homozygous recessive (rr) genotype results in a colorless one. The grains will appear either purple or yellow rather than purple or colorless as you might expect. The yellow is due to the presence of yellow endosperm below the colorless, transparent aleurone layer. The allele Su produces starchy endosperm and the allele su produces sweet endosperm. Su is completely dominant over su. When the kernels dry, SuSu and Susu grains are smooth, while susu grains are wrinkled. Predict what phenotypic ratio of purple-smooth; purple-wrinkled; yellow-smooth; yellow-wrinkled kernels you expect to observe. 37. What is your expected dihybrid F2 phenotypic ratio. 38. What is your null hypothesis? I tol 39. What is your alternative hypothesis? 71
Transcribed Image Text:34. What are the possible genotypes of the progeny?. What are the phenotypes of the progeny?. c. The F1 plants of problem b above are self-pollinated. (For this problem, assume that no crossing-over has taken place.) 35. What alleles will their gametes carry? eggs 36. What are the genotypes of the F2 population?. What are the phenotypes of the F2 population?_ sperm Part B Maize Genetics Aleurone Color and Endosperm Characteristics In the corn grain, the aleurone layer is a single layer of cells between the pericarp and the endosperm. The color of the aleurone layer is determined by several interacting genes, one of which, R, is responsible for anthocyanin production in the aleurone layer. Homozygous dominant (RR) and heterozygous (Rr) genotypes result in a purple aleurone layer and the homozygous recessive (rr) genotype results in a colorless one. The grains will appear either purple or yellow rather than purple or colorless as you might expect. The yellow is due to the presence of yellow endosperm below the colorless, transparent aleurone layer. The allele Su produces starchy endosperm and the allele su produces sweet endosperm. Su is completely dominant over su. When the kernels dry, SuSu and Susu grains are smooth, while susu grains are wrinkled. Predict what phenotypic ratio of purple-smooth; purple-wrinkled; yellow-smooth; yellow-wrinkled kernels you expect to observe. 37. What is your expected dihybrid F2 phenotypic ratio. 38. What is your null hypothesis? I tol 39. What is your alternative hypothesis? 71
Dihybrid Cross: Examine an ear of corn representing an F2 generation. Do not remove any grains from the ears of corn, do not remove the plastic sleeves that cover the ears, and please handle the ears carefully. If you wish, you can use a very small piece of masking tape at one of the ear to keep track of the rows as you count the grains. Do not count all of the purple and then return to count all of the yellow. This will lead to some grains being counted twice and some not being counted at all. Rather, examine each grain one at a time and determine if it is purple or yellow and if it is wrinkled or smooth. Work with your lab partner. One person can call out the phenotype while the other person records the phenotype. If you are unsure how to categorize particular grains, you may consult your lab instructor. 40. Record your results below. Number of purple, smooth kernels 5 Number of purple, wrinkled kernels Number of yellow, smooth kernels Number of yellow, wrinkled kernels. 118 41. What are the phenotypes of the F2 generation? Of the F1 generation. 42. What are the genotypes of the F2 generation? Of the generation 43. How do the observed phenotypes of your ear of corn compare with the expected F2 ratio? To calculate your expected ratio numbers given the total number of corn kernels that you counted use the following steps: A. Determine your expected ratio. For a dihybrid cross between two F₁ plants, both heterozygous for both traits, you would expect a 9:3:3:1 ratio. If you add 9+3+3+1, it equals 16. Therefore, each category is a fraction of 16. This means that you expect the 2 following fractions 9/16, 3/16, 3/16, 1/16 for each phenotypic class. B. Determine your total number of corn kernels actually counted. C. Multiple each fraction by the total number of corn kernels actually counted. a. For example, if you counted 215 kernels, you would expect the following: i. 9/16 X 215 = 121 dominant phenotype for both traits ii. 3/16 X 215 = 40 dominant phenotype for 1 trait and recessive phenotype for other trait iii. 3/16 X 215 = 40 dominant phenotype for 1 trait and recessive phenotype for other trait iv. 1/16 X 215 = 14 recessive phenotype for both traits o D. Add the 4 resulting numbers to verify that they add up to your total number E. Compare the 4 resulting numbers (your expected numbers) to your actual numbers using Chi Square statistical analysis (to be completed later in Exercise III). 72
Transcribed Image Text:Dihybrid Cross: Examine an ear of corn representing an F2 generation. Do not remove any grains from the ears of corn, do not remove the plastic sleeves that cover the ears, and please handle the ears carefully. If you wish, you can use a very small piece of masking tape at one of the ear to keep track of the rows as you count the grains. Do not count all of the purple and then return to count all of the yellow. This will lead to some grains being counted twice and some not being counted at all. Rather, examine each grain one at a time and determine if it is purple or yellow and if it is wrinkled or smooth. Work with your lab partner. One person can call out the phenotype while the other person records the phenotype. If you are unsure how to categorize particular grains, you may consult your lab instructor. 40. Record your results below. Number of purple, smooth kernels 5 Number of purple, wrinkled kernels Number of yellow, smooth kernels Number of yellow, wrinkled kernels. 118 41. What are the phenotypes of the F2 generation? Of the F1 generation. 42. What are the genotypes of the F2 generation? Of the generation 43. How do the observed phenotypes of your ear of corn compare with the expected F2 ratio? To calculate your expected ratio numbers given the total number of corn kernels that you counted use the following steps: A. Determine your expected ratio. For a dihybrid cross between two F₁ plants, both heterozygous for both traits, you would expect a 9:3:3:1 ratio. If you add 9+3+3+1, it equals 16. Therefore, each category is a fraction of 16. This means that you expect the 2 following fractions 9/16, 3/16, 3/16, 1/16 for each phenotypic class. B. Determine your total number of corn kernels actually counted. C. Multiple each fraction by the total number of corn kernels actually counted. a. For example, if you counted 215 kernels, you would expect the following: i. 9/16 X 215 = 121 dominant phenotype for both traits ii. 3/16 X 215 = 40 dominant phenotype for 1 trait and recessive phenotype for other trait iii. 3/16 X 215 = 40 dominant phenotype for 1 trait and recessive phenotype for other trait iv. 1/16 X 215 = 14 recessive phenotype for both traits o D. Add the 4 resulting numbers to verify that they add up to your total number E. Compare the 4 resulting numbers (your expected numbers) to your actual numbers using Chi Square statistical analysis (to be completed later in Exercise III). 72
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