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 t 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.
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 t 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.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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
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.

Transcribed Image Text:C) Based on the p value you determined, should you reject the null hypothesis?
o reject; the difference is due to chance
do not reject; the difference is due to chance
do not reject; the difference is not due to chance
o reject; the difference is not due to chance
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