2) Another new mutation for yellow eyes is thought to be X- linked recessive. The P1 virgin females were mated with yellow-eyed males. 50 wild type females and 50 wild type males of the F1 generation are bred. The resulting offspring are: no yellow eyed females, 33 yellow-eyed males, 80 wild type females, and 35 wild - type males. a. What is your null hypothesis? (Hint: the foundation of this is always the same, that there is no statistical difference between your observed data and the expected data) <3 37 14 37 37 . 25 b. Do a chi-square test to either accept or reject your null hypothesis. c. Explain why your null hypothesis is either accepted or rejected. 35 37

Females   XAXA  X  XaY (Yellow male)

F1         XAXa (WT Female) : XAY (WT Male)

F2         XAXA, XAXa (WT Female) : XAY (WT Male) : XaY (Yellow male)

Ratio                    2        :      1        :      1

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A. Null Hypothesis H0 - The mutation follows X-linked Recessive pattern with ratio 1:2:1

B. Accept the Null hypothesis.

C. Chisquare value = 1.027                        

Degree of freedom n-1 = 3-1 = 2                        

P0.05 = 5.991                        

Chisquare value = 1.207 < P0.05                        

So, Accept the Null hypothesis.                        

There were no significative differences, by the chi-square test at the level of 5% of probability,            

between the observed and expected ratios. So they are consistent with the ratio.    

Please see the attached file for the complete solution