Chi-squared Problems 1. Homozygous green peas were crossed with homozygous yellow peas. All the offspring were phenotypically yellow. A test cross was performed with one of the offspring from the F1. A. Using Y as the symbol for pea color, what are the genotypes of the parents used in this cross? B. If the F2 generation consisted of 20 offspring, predict how many will be green and how many will be yellow. C. If the offspring consisted of eight yellow peas and twelve green peas, would this be statistically different from your prediction? Use a Chi-squared test on the F2 generation data to analyze the results of the cross. Show all your work. What is the best explanation for the difference between what was expected and what was observed? D. If there were 200 offspring and they consisted of 80 yellow peas and 120 green peas, would this be statistically different from your prediction? Use a Chi-squared test on the F2 generation data to analyze the results of the cross. Show all your work. 2. In fruit flies, the phenotype for eye color is determined by a certain locus. E indicates the dominant allele and e indicates the recessive allele. The cross between a male wild-type fruit fly and a female white-eyed fruit fly produced the following offspring. F1 Generation Wild Type Male Wild Type Female Whiteeyed Male White-eyed Female Brown-eyed Female 0 45 55 0 1 The wild-type and white-eyed individuals from the F1 generation were then crossed to produce the following offspring. F2 Generation Wild Type Male Wild Type Female Whiteeyed Male White-eyed Female Brown-eyed Female 23 31 22 24 0 A. Determine the genotypes of the original parents (P generation) and explain your reasoning. You may use punnett squares to enhance your description, but the results from the punnett squares must be discussed in your answer. B. Use a Chi-squared test on the F2 generation data to analyze your prediction of the parental genotypes. Show all your work and explain the importance of your final answer. C. The brown-eyed female in the F1 generation resulted from a mutational change. Explain what a mutation is, and discuss two types of mutations that might have produced the brown-eyed female in the F1 generation. 3. In certain strains of corn, the phenotype for color can be yellow or purple. Also the texture of the kernels can be “plump” (smooth) or the kernels can be wrinkled. The cross between a homozygous purple and wrinkled strain of corn was crossed with a homozygous yellow and smooth strain of corn. The offspring from this cross produced were all smooth and purple. The F1 offspring were crossed with one another and the phenotypes of the kernels from several ears of corn were counted and recorded in the data table below for the F2 generation. F2 Generation Purple, Smooth Purple Wrinkled Yellow Smooth Yellow Wrinkled 375 133 118 47 A. What is the null hypothesis for this investigation? B. Use a Chi-squared test on the F2 generation data to analyze the results of the cross to determine if the results supports that these genes are unlinked. Show all your work and explain the importance of your final answer. 4. In an investigation of pill bug behavior, a covered choice chamber is used to test whether the distribution of pill bugs is affected by the presence of a sliced potato. A slice of potato is placed at one end of the chamber and dry cotton is placed at the other end. To test the pill bugs preference for the potato 25 pill bugs are placed on each side of the choice chamber. The positions of pill bugs are observed and recorded every 10 minutes for 30 minutes. Time (minutes) 0 10 20 30 Side with Potato 25 21 17 14 Side with Cotton 25 29 33 36 A. What is the null hypothesis for this investigation? B. Perform a chi-square test on the data for the last data collected at the 30-minute time point in the investigation. Explain whether your hypothesis is supported by the chi-square test and justify your explanation. 5. Onion roots were treated with 1 M solution of caffeine, and other onion roots were just submerged in water with no treatment. After three days, four samples of the roots were removed and a root tip squash was done. The tips were observed under the microscope. The total number cells observed were counted and the number of cells undergoing of mitosis were also counted. Root tip with no treatment Phase Count Root tip treated with caffeine Phase Count Interphase 212 Interphase 370 Mitosis 72 Mitosis 51 Total Cell Count 284 Total Cell Count 421 % Cell in Mitosis % Cells in Mitosis A. What is the null hypothesis for this investigation? B. Perform a chi-square test on the data for the data collected caffeine treatment t in the investigation. Explain whether your hypothesis is supported by the chi-square test and justify your explanation. 6. In a field of flowers, there were three phenotypic colors for a particular species purple, violet and white. It was determined that the color was controlled by a single autosomal gene that exhibited partial dominance. The following counts were made: Number Purple Flowers PP 243 Violet Flowers PP' 440 A. Determine the allelic frequency for the P and P’ alleles. White Flowers P'P' 36 Number Purple Flowers PP Violet Flowers PP' White Flowers P'P" Counts #P Alleles X 243 440 36 # P' Alleles X Grand Total Alleles Total Allelic Frequency B. Determine if this field was in Hardy-Weinberg equilibrium using the chi-square test. Chi-squared Problems Key 1. Homozygous green peas were crossed with homozygous yellow peas. All the offspring were phenotypically yellow. A test cross was performed with one of the offspring from the F1. A. Using Y as the symbol for pea color, what are the genotypes of the parents used in this cross? YY x yy B. If the F2 generation consisted of 20 offspring, predict how many will be green and how many will be yellow. One half will be yellow and one half will green as the cross is between a green pea plant that is heterozygous Yy and the other is homozygous for green yy. The results should be ten that are green and ten that are yellow y y Y Yy Yy y yy yy C. If the offspring consisted of eight yellow peas and twelve green peas, would this be statistically different from your prediction? Use a Chi-squared test on the F2 generation data to analyze the results of the cross. Show all your work. What is the best explanation for the difference between what was expected and what was observed? ∑ (12-10)2 (8-10)2 + =0.8 0.8 < 3.84 which means that the difference is not statistically 10 10 significant. The difference can be accounted for by the small sample size. D. If there were 200 offspring and they consisted of 80 yellow peas and 120 green peas, would this be statistically different from your prediction? Use a Chi-squared test on the F2 generation data to analyze the results of the cross. Show all your work. ∑ (120-100)2 (80-100)2 + =8 8 > 3.84 which means that the difference is statistically 100 100 significant. The results do not support that this cross is a simple Mendelian cross controlled by one simple gene. 2. In fruit flies, the phenotype for eye color is determined by a certain locus. E indicates the dominant allele and e indicates the recessive allele. The cross between a male wild-type fruit fly and a female white-eyed fruit fly produced the following offspring. F1 Generation Wild Type Male Wild Type Female Whiteeyed Male White-eyed Female Brown-eyed Female 0 45 55 0 1 The wild-type and white-eyed individuals from the F1 generation were then crossed to produce the following offspring. F2 Generation Wild Type Male Wild Type Female Whiteeyed Male White-eyed Female Brown-eyed Female 23 31 22 24 0 A. Determine the genotypes of the original parents (P generation) and explain your reasoning. You may use punnett squares to enhance your description, but the results from the punnett squares must be discussed in your answer. This is a sex linked cross with the white being recessive to wild type. The genotypes of the parents involved in the P cross are XeXe x XEY. All the female offspring are phenotypically red and all the male offspring are phenotypically white. XE Y Xe XEXe Xey Xe XEXe Xey B. Use a Chi-squared test on the F2 generation data to analyze your prediction of the parental genotypes. Show all your work and explain the importance of your final answer. The parental genotypes are XeXe x XEY and based on the punnett square, there should 25% of the total for each phenotype. Xe Y XE XEXe XEy Xe XeXe Xey Wild Type Male χ2 = Wild White- White- BrownType eyed eyed eyed Female Male Female Female Total F2 Generation Observed 23 31 22 24 Expected 25 25 25 25 (23 − 25)2 (31 − 25)2 (22 − 25)2 (24 − 25)2 + + + 25 25 25 25 0 100 =2 2<7.82 the differences between what was expected and what was observed is not statistically significant. C. The brown-eyed female in the F1 generation resulted from a mutational change. Explain what a mutation is, and discuss two types of mutations that might have produced the brown-eyed female in the F1 generation. This may be the result of a mutation in the production of the pigment such as a point mutation or it may be the result of a mutation in the biochemical pathway responsible for the deposition of the pigment. 3. In certain strains of corn, the phenotype for color can be yellow or purple. Also the texture of the kernels can be “plump” (smooth) or the kernels can be wrinkled. The cross between a homozygous purple and wrinkled strain of corn was crossed with a homozygous yellow and smooth strain of corn. The offspring from this cross produced were all smooth and purple. The F1 offspring were crossed with one another and the phenotypes of the kernels from several ears of corn were counted and recorded in the data table below for the F2 generation. F2 Generation Purple, Smooth Purple Wrinkled Yellow Smooth Yellow Wrinkled 375 133 118 47 A. What is the null hypothesis for this investigation? There is no statistical difference between the results of this cross and the predicted 9:3:3:1 ratio for a Mendelian dihybrid cross. B. Use a Chi-squared test on the F2 generation data to analyze the results of the cross to determine if the results supports that these genes are unlinked. Show all your work and explain the importance of your final answer. F2 Generation Expected Expected Difference χ2 = Purple, Smooth Purple Wrinkled Yellow Smooth Yellow Wrinkled 375 133 118 47 9/16X(673) 3/16X(673) 3/16X(673) 379 126 126 4 -7 8 (375 − 379)2 (133 − 126)2 (118 − 126)2 (47 − 42)2 + + + 379 126 126 42 Total 673 1/16 X(673) 42 -5 =1.5 1.5<7.82 the differences between what was expected and what was observed is not statistically significant. 4. In an investigation of pill bug behavior, a covered choice chamber is used to test whether the distribution of pill bugs is affected by the presence of a sliced potato. A slice of potato is placed at one end of the chamber and dry cotton is placed at the other end. To test the pill bugs preference for the potato 25 pill bugs are placed on each side of the choice chamber. The positions of pill bugs are observed and recorded every 10 minutes for 30 minutes. Time (minutes) 0 10 20 30 Side with Potato 25 21 17 14 Side with Cotton 25 29 33 36 A. What is the null hypothesis for this investigation? Providing pill bugs with the potato has no effect on the distribution of the pill bugs in the choice chambers. B. Perform a chi-square test on the data for the last data collected at the 30-minute time point in the investigation. Explain whether your hypothesis is supported by the chi-square test and justify your explanation. χ = 2 (14 − 25)2 (36 − 25)2 =9.68 9.68>3.84 The data supports that the difference between what + 25 25 was observed and what was expected is statistically significant. The data supports that providing pill bugs potato has an effect on their distribution. 5. Onion roots were treated with 1 M solution of caffeine, and other onion roots were just submerged in water with no treatment. After three days, four samples of the roots were removed and a root tip squash was done. The tips were observed under the microscope. The total number cells observed were counted and the number of cells undergoing of mitosis were also counted. Root tip with no treatment Phase Count Root tip treated with caffeine Phase Count Interphase 212 Interphase 370 Mitosis 72 Mitosis 51 Total Cell Count 284 Total Cell Count 421 % Cells in Mitosis % Cells in Interphase 0.25 0.75 A. What is the null hypothesis for this investigation? Caffeine has no effect on the mitotic division of cells in the root tip of an onion. B. Perform a chi-square test on the data for the data collected caffeine treatment t in the investigation. Explain whether your hypothesis is supported by the chi-square test and justify your explanation. χ = 2 Root tip treated with caffeine # cells in mitosis # of cells in interphase Observed 51 370 Expected Expected Difference 421 x .25 105.25 -54.25 421 x .75 315.75 54.25 (51 − 105)2 (370 − 316)2 =37.28 37.28>3.84 which supports that the difference between what + 105 316 was expected and what was observed is significant. This supports that a 1 M solution of caffeine has a negative effect on the mitosis of onion root tips. 6. In a field of flowers, there were three phenotypic colors for a particular species purple, violet and white. It was determined that the color was controlled by a single autosomal gene that exhibits partial dominance. The following counts were made: Number Purple Flowers PP 243 Violet Flowers PP' 440 White Flowers P'P' 36 A. Determine the allelic frequency for the P and P’ alleles using the data provided in the table shown. Number Purple Flowers PP Violet Flowers PP' White Flowers P'P" Counts #P Alleles # P' Alleles 243 486 X 440 440 440 36 X 72 Grand Total Alleles 512 1438 Total 926 Allelic Frequency B. Determine if these flowers found in the field was in Hardy-Weinberg equilibrium by using the chi-square test. If the flowers in the field is in Hardy-Weinber equ