PART III 9 will perform a testcross for flies of a different

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 6

Solution

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