PART III 9. will perform a testcross for flies of a different mutant line. You h
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Question
PART III 9. will perform a testcross for flies of a different mutant line. You have established a true-breeding mutant line with curved wings (c), which is controlled by a single gene on chromosome 2, and purple eyes which is controlled by a single gene on chromosome 2. You establish a parental cross between true- breeding males with curved wings and purple eyes and true-breeding wild-type females. You then mate the F1 females to males that are true-breeding for cross curved wings and purple eyes to generate the F2 generation. Map out this at all below, including expected genotypes and phenotypes of males and females stages of the experiment. Include the expected phenotypic ratio for the F2 generation. 10. using a virtual fly experiment a6, and the results are provided below: we completed the cross described in Number of Number of Total females males observed 1233 F1 phenotypes wild-type wings eyes wild-type wings purple eyes curved wings wild-type eyes curved wings purple eyes Number of Number of Total females males observed 240 470 observed F2 phenotypes wild-type wings wild-type eyes wild-type wings purple eyes 43 45 115 curved wings wild-type eyes 244 249 493 purple eyes dihybrid testcross these consistent with the expected outcomes of a provide support assuming independent assortment? Use a chi-square analysis to for your answer show 5 of 6Explanation / Answer
a)
Null hypthoesis:
The observed values are not deviating from the 9:3:3:1 ratio.
Categories
Wild Wild
Wild purple
Curved Wild
Curved purple
Total
Observed Values
470
88
115
493
1166
Expected ratio
9
3
3
1
16
Exprected values
655.875
218.625
218.625
72.875
Deviation
-185.875
-130.625
-103.625
420.125
D^2
34549.51563
17062.891
10738.14063
176505.0156
D^2/E
52.67698209
78.046384
49.11670955
2422.024228
2601.864
X^2
2601.864303
Degrees of Freedom
No of categories - 1
4-1
3
b)
Exprected values
655.875 218.625 218.625 72.875
c)
What is your calculted chisquare test statistic
2601.864303
D)
Degrees of freedom
No of categories - 1: 4-1: 3
E)
Inference: The caculated chisquare value is greater than the table value i.e. 7.82 at 0.05 probablity and 3 DF. Hense Null hypothesis rejected. This indicates that the results are indicating these two genes are a linked genes.
Categories
Wild Wild
Wild purple
Curved Wild
Curved purple
Total
Observed Values
470
88
115
493
1166
Expected ratio
9
3
3
1
16
Exprected values
655.875
218.625
218.625
72.875
Deviation
-185.875
-130.625
-103.625
420.125
D^2
34549.51563
17062.891
10738.14063
176505.0156
D^2/E
52.67698209
78.046384
49.11670955
2422.024228
2601.864
X^2
2601.864303
Degrees of Freedom
No of categories - 1
4-1
3
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