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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 test the null hypothesis.

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 test the null 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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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.
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.
df .995 .975 .9 .5 .1 .05 .025 .01 .005 .000 .000 0.016 0.455 2.706 3.841 5.024 6.635 7.879 B) Using the chi square table, the range of p values based on this data is between: 0.010 0.051 0.211 1.386 4.605 5.991 7.378 9.210 10.597 3 0.072 0.216 0.584 2.366 6.251 7.815 9.348 11.345 12.838 4 0.207 0.484 1.064 3.357 7.779 9.488 11.143 13.277 14.860 o .05 and .1 o .1 and .5 O .5 and .9 o .025 and .05 o .01 and .025 O .9 and .975
Transcribed Image Text:df .995 .975 .9 .5 .1 .05 .025 .01 .005 .000 .000 0.016 0.455 2.706 3.841 5.024 6.635 7.879 B) Using the chi square table, the range of p values based on this data is between: 0.010 0.051 0.211 1.386 4.605 5.991 7.378 9.210 10.597 3 0.072 0.216 0.584 2.366 6.251 7.815 9.348 11.345 12.838 4 0.207 0.484 1.064 3.357 7.779 9.488 11.143 13.277 14.860 o .05 and .1 o .1 and .5 O .5 and .9 o .025 and .05 o .01 and .025 O .9 and .975
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