Chi Square Genetics 2022

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102

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Biology

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Jan 9, 2024

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Name:___________________________________________________________________ Period:_________________ Chi Square Genetics Chi-square is a statistical tool that helps us to decide if the observed ratio is close enough to the expected ratio to acceptable. Chi-square analysis can be used in genetics crosses. Whenever you have to determine if an expected ratio fits an observed ratio, you can use Chi-square! 1. In peas, yellow seeds (A) are dominant over green seeds (a). In a cross between two plants both heterozygous for seed color, the following was observed: yellow = 4400 green = 1624 What do you predict the expected F2 phenotypic ratio to be? Describe the MODE OF INHERITANCE for this cross: (circle answer for each) Is the trait sex-linked or autosomal? Is the trait dominant or recessive? State a NULL hypothesis with the expected phenotypic ratio of this experiment: Determine the Chi Square value: Phenotype Observed (o) Expected (e) (o-e) (o-e) 2 (o-e) 2 e c 2 = S (o-e) 2 e Degrees of freedom (df) = ______ What is the Critical value (get from equation sheet!) = _________________ Conclusion : Do you support or not support the null hypothesis? Why?

2. In peas, smooth seeds (A) are dominant over wrinkled (a) seeds. In the P generation, a plant heterozygous for smooth seeds is crossed with a plant with wrinkled seeds. Then resulting F1 plants are crossed. The seeds of the observed F2 generation were: smooth = 5474 wrinkled = 1850 What do you predict the expected F2 phenotypic ratio to be? Describe the MODE OF INHERITANCE for this cross: Is the trait sex-linked or autosomal? Is the trait dominant or recessive? State a NULL hypothesis with the expected ratio for the experiment: Calculate Chi Square Value: Phenotype Observed (o) Expected (e) (o-e) (o-e) 2 (o-e) 2 e c 2 = S (o-e) 2 e Degrees of freedom (df) = _____________ Critical Value = _____________ Conclusion : Do you support or not support the null hypothesis? Why?

3. In a flowering plant, white flowers (A) are dominant over red (a), and short plants (B) are dominant over tall (b) plants. When two double heterozygote (AaBb) plants were crossed, the resulting phenotypes were observed: white, short = 206 red, short = 83 white, tall = 65 red, tall = 30 What do you predict the expected phenotypic ratio to be? Describe the MODE OF INHERITANCE for this cross: Is the trait sex-linked or autosomal? Is the trait dominant or recessive? State a NULL hypothesis with expected ratio for this experiment: Calculate Chi Square Value Phenotype Observed (o) Expected (e) (o-e) (o-e) 2 (o-e) 2 e c 2 = S (o-e) 2 e Degrees of freedom (df) = Critical Value = Conclusion : Do you support or not support the null hypothesis? Why?

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4. An investigator observes that when pure-breeding, long-winged fruit flies are mated with pure-breeding, short-winged flies, the F 1 offspring have an intermediate wing length. When several intermediate-winged flies are allowed to interbreed, the following results are obtained: 230 long-winged flies 510 intermediate winged flies 260 short winged flies. What do you predict the expected F2 phenotypic ratio to be? Describe the MODE OF INHERITANCE for this cross: Is the trait sex-linked or autosomal? Is the trait dominant or recessive? State a NULL hypothesis for this experiment: Calculate the Chi Square Value: Phenotype Observed (o) Expected (e) (o-e) (o-e) 2 (o-e) 2 e c 2 = S (o-e) 2 e Degrees of freedom (df)= Critical value = Conclusion : Do you support or not support the null hypothesis? Why?

5. Color blindness is a sex-linked trait in Wombats. A female who is a carrier (not colorblind) mates with a male who is color blind. The phenotypes of their offspring are: Normal female = 132 Color blind female = 124 Normal male = 126 Color blind male = 136 What do you predict the expected F2 phenotypic ratio to be? Describe the MODE OF INHERITANCE for this cross: Is the trait sex-linked or autosomal? Is the trait dominant or recessive? State a NULL hypothesis for this experiment: Calculate the Chi Square Value: Phenotype Observed (o) Expected (e) (o-e) (o-e) 2 (o-e) 2 e c 2 = S (o-e) 2 e Degrees of freedom (df)= Critical value = Conclusion : Do you support or not support the null hypothesis? Why?

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