Short Answer 5. You identify two recessive mutations in fruit flies, one that results in curly wings (cc) and one that results in an ebony body (ee). You cross two flies that are heterozygous for both traits (CcEe), and find 352 flies progeny in the next generation with the following phenotypes: wild type - 193 wild type body, curly wings - 64 wild type wings, ebony body - 69 curly wings, ebony body - 26 You believe the curly and ebony alleles assort independently and are not linked. Perform a chi square analysis for this cross to test the null hypothesis. A) Show the expected numbers for each phenotype, and the final X value. Observed Expected (0-e)?/e x2 Edit View Insert Format Tools Table

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
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Short Answer 5. You identify two recessive mutations in fruit flies, one that results in curly wings (cc) and one that results in an ebony body (ee). You cross two flies that are heterozygous for both traits (CCEE), and find 352 flies progeny in the next generation with the
following phenotypes:
wild type – 193
wild type body, curly wings - 64
wild type wings, ebony body – 69
curly wings, ebony body - 26
You believe the curly and ebony alleles assort independently and are not linked. Perform a chi square analysis for this cross to test the null hypothesis.
A) Show the expected numbers for each phenotype, and the final X value.
Observed Expected (o-e)²/e
Edit View Insert Format Tools Table
Transcribed Image Text:Short Answer 5. You identify two recessive mutations in fruit flies, one that results in curly wings (cc) and one that results in an ebony body (ee). You cross two flies that are heterozygous for both traits (CCEE), and find 352 flies progeny in the next generation with the following phenotypes: wild type – 193 wild type body, curly wings - 64 wild type wings, ebony body – 69 curly wings, ebony body - 26 You believe the curly and ebony alleles assort independently and are not linked. Perform a chi square analysis for this cross to test the null hypothesis. A) Show the expected numbers for each phenotype, and the final X value. Observed Expected (o-e)²/e Edit View Insert Format Tools Table
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