Your goal is to produce black seeds resistant to mold. So you make the same cross again (between a homozygous black seeded, mold susceptible parent and a homozygous white seeded and mold resistant parent), and, again, advance progeny by SSD to create 100 F10 generation plants. Based on the information you obtained from your first crossing experiment (Question #4), how many F10 plants would you expect to have black seeds and be resistant to mold? Assume that a toxin produced by the mold fungus has been isolated. Only mold resistant seeds will germinate in the presence of the toxin. Could you use this toxin screening procedure to have segregation distortion work in your favor in the F2 generation? Explain your answer. Info from Question 4 a. P Locus (Seed Color): Hypothesis: The null hypothesis (H₀) is that seed color is controlled by alleles at a single locus. Observed Data: Total white seeds: 45 (resistant plants) + 6 (susceptible plants) = 51 Total black seeds: 7 (resistant plants) + 42 (susceptible plants) = 49 Total seeds: 51 + 49 = 100 Expected Data (1:1 ratio for a single locus): Expected white seeds: 100 / 2 = 50 Expected black seeds: 100 / 2 = 50 Chi-Square Calculation: χ² = Σ [(observed - expected)² / expected] χ² = [(51 - 50)² / 50] + [(49 - 50)² / 50] χ² = (1² / 50) + (-1² / 50) χ² = (1 / 50) + (1 / 50) χ² = 2 / 50 = 0.04 Degrees of Freedom (df) = number of categories - 1 = 2 - 1 = 1 P-Value: Compare the calculated χ² value (0.04) with a chi-square distribution table for df = 1. A χ² value of 0.04 corresponds to a p-value greater than 0.05. Since the p-value is greater than 0.05, we fail to reject the null hypothesis. This shows that the observed data is not significantly different from the expected 1:1 ratio. Meaning that seed color is likely controlled by alleles at a single locus. M Locus (Mold Resistance): Hypothesis: The null hypothesis (H0) is that mold resistance is controlled by alleles at a single locus. Observed Data: Resistant plants: 52 Susceptible plants: 48 Total plants: 52 + 48 = 100 Expected Data (1:1 ratio for a single locus): Expected resistant plants: 100 / 2 = 50 Expected susceptible plants: 100 / 2 = 50 Chi-Square Calculation: χ² = Σ [(observed - expected)² / expected] χ² = [(52 - 50)² / 50] + [(48 - 50)² / 50] χ² = (2² / 50) + (-2² / 50) χ² = (4 / 50) + (4 / 50) χ² = 8 / 50 = 0.16 Degrees of Freedom (df) = number of categories - 1 = 2 - 1 = 1 P-Value: Compare the calculated χ² value (0.16) with a chi-square distribution table for df = 1. A χ² of 0.16 corresponds to a p-value greater than 0.05. Since the p-value is greater than 0.05, we fail to reject the null hypothesis. So the observed data is not significantly different from the expected 1:1 ratio. We can conclude that mold resistance is likely controlled by alleles at a single locus. In both cases, the chi-square tests show that the observed phenotypic ratios for seed color and mold resistance are consistent with a single-locus inheritance pattern. The high p-values suggest that any deviations from the expected 1:1 ratios are probably due to random chance. We can confidently conclude that each trait is likely controlled by alleles at a single locus.

Your goal is to produce black seeds resistant to mold. So you make the same cross again (between a homozygous black seeded, mold susceptible parent and a homozygous white seeded and mold resistant parent), and, again, advance progeny by SSD to create 100 F10 generation plants. Based on the information you obtained from your first crossing experiment (Question #4), how many F10 plants would you expect to have black seeds and be resistant to mold? Assume that a toxin produced by the mold fungus has been isolated. Only mold resistant seeds will germinate in the presence of the toxin. Could you use this toxin screening procedure to have segregation distortion work in your favor in the F2 generation? Explain your answer. Info from Question 4 a. P Locus (Seed Color): Hypothesis: The null hypothesis (H₀) is that seed color is controlled by alleles at a single locus. Observed Data: Total white seeds: 45 (resistant plants) + 6 (susceptible plants) = 51 Total black seeds: 7 (resistant plants) + 42 (susceptible plants) = 49 Total seeds: 51 + 49 = 100 Expected Data (1:1 ratio for a single locus): Expected white seeds: 100 / 2 = 50 Expected black seeds: 100 / 2 = 50 Chi-Square Calculation: χ² = Σ [(observed - expected)² / expected] χ² = [(51 - 50)² / 50] + [(49 - 50)² / 50] χ² = (1² / 50) + (-1² / 50) χ² = (1 / 50) + (1 / 50) χ² = 2 / 50 = 0.04 Degrees of Freedom (df) = number of categories - 1 = 2 - 1 = 1 P-Value: Compare the calculated χ² value (0.04) with a chi-square distribution table for df = 1. A χ² value of 0.04 corresponds to a p-value greater than 0.05. Since the p-value is greater than 0.05, we fail to reject the null hypothesis. This shows that the observed data is not significantly different from the expected 1:1 ratio. Meaning that seed color is likely controlled by alleles at a single locus. M Locus (Mold Resistance): Hypothesis: The null hypothesis (H0) is that mold resistance is controlled by alleles at a single locus. Observed Data: Resistant plants: 52 Susceptible plants: 48 Total plants: 52 + 48 = 100 Expected Data (1:1 ratio for a single locus): Expected resistant plants: 100 / 2 = 50 Expected susceptible plants: 100 / 2 = 50 Chi-Square Calculation: χ² = Σ [(observed - expected)² / expected] χ² = [(52 - 50)² / 50] + [(48 - 50)² / 50] χ² = (2² / 50) + (-2² / 50) χ² = (4 / 50) + (4 / 50) χ² = 8 / 50 = 0.16 Degrees of Freedom (df) = number of categories - 1 = 2 - 1 = 1 P-Value: Compare the calculated χ² value (0.16) with a chi-square distribution table for df = 1. A χ² of 0.16 corresponds to a p-value greater than 0.05. Since the p-value is greater than 0.05, we fail to reject the null hypothesis. So the observed data is not significantly different from the expected 1:1 ratio. We can conclude that mold resistance is likely controlled by alleles at a single locus. In both cases, the chi-square tests show that the observed phenotypic ratios for seed color and mold resistance are consistent with a single-locus inheritance pattern. The high p-values suggest that any deviations from the expected 1:1 ratios are probably due to random chance. We can confidently conclude that each trait is likely controlled by alleles at a single locus.

Biology (MindTap Course List)

11th Edition

ISBN:9781337392938

Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Publisher:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Chapter11: The Basic Principles Of Heredity

Section: Chapter Questions

Problem 11TYU: Individuals of genotype AaBb were mated to individuals of genotype aabb. One thousand offspring were...

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Based on the information you obtained from your first crossing experiment (Question #4), how many F10 plants would you expect to have black seeds and be resistant to mold?
Assume that a toxin produced by the mold fungus has been isolated. Only mold resistant seeds will germinate in the presence of the toxin. Could you use this toxin screening procedure to have segregation distortion work in your favor in the F2 generation? Explain your answer.
Info from Question 4
a. P Locus (Seed Color):

Hypothesis: The null hypothesis (H₀) is that seed color is controlled by alleles at a single locus.
Observed Data:

Total white seeds: 45 (resistant plants) + 6 (susceptible plants) = 51
Total black seeds: 7 (resistant plants) + 42 (susceptible plants) = 49
Total seeds: 51 + 49 = 100

Expected Data (1:1 ratio for a single locus):

Expected white seeds: 100 / 2 = 50
Expected black seeds: 100 / 2 = 50

Chi-Square Calculation:

χ² = Σ [(observed - expected)² / expected]
χ² = [(51 - 50)² / 50] + [(49 - 50)² / 50]
χ² = (1² / 50) + (-1² / 50)
χ² = (1 / 50) + (1 / 50)
χ² = 2 / 50 = 0.04

Degrees of Freedom (df) = number of categories - 1 = 2 - 1 = 1
P-Value: Compare the calculated χ² value (0.04) with a chi-square distribution table for df = 1.

A χ² value of 0.04 corresponds to a p-value greater than 0.05.

Since the p-value is greater than 0.05, we fail to reject the null hypothesis. This shows that the observed data is not significantly different from the expected 1:1 ratio. Meaning that seed color is likely controlled by alleles at a single locus.

M Locus (Mold Resistance):

Hypothesis: The null hypothesis (H₀) is that mold resistance is controlled by alleles at a single locus.
Observed Data:

Resistant plants: 52
Susceptible plants: 48
Total plants: 52 + 48 = 100

Expected Data (1:1 ratio for a single locus):

Expected resistant plants: 100 / 2 = 50
Expected susceptible plants: 100 / 2 = 50

Chi-Square Calculation:

χ² = Σ [(observed - expected)² / expected]
χ² = [(52 - 50)² / 50] + [(48 - 50)² / 50]
χ² = (2² / 50) + (-2² / 50)
χ² = (4 / 50) + (4 / 50)
χ² = 8 / 50 = 0.16

Degrees of Freedom (df) = number of categories - 1 = 2 - 1 = 1
P-Value: Compare the calculated χ² value (0.16) with a chi-square distribution table for df = 1.

A χ² of 0.16 corresponds to a p-value greater than 0.05.

Since the p-value is greater than 0.05, we fail to reject the null hypothesis. So the observed data is not significantly different from the expected 1:1 ratio. We can conclude that mold resistance is likely controlled by alleles at a single locus.

In both cases, the chi-square tests show that the observed phenotypic ratios for seed color and mold resistance are consistent with a single-locus inheritance pattern. The high p-values suggest that any deviations from the expected 1:1 ratios are probably due to random chance. We can confidently conclude that each trait is likely controlled by alleles at a single locus.

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