In flies dumpy wings and ebony bodies are each mutations recessive to wild type, and you expect them to behave in a Mendelian fashion. A dihybrid cross between two animals, each heterozygous for dumpy wings and ebony bodies (DdEe X DdEe) is performed with the following results:

Wild type wings; wild type body 473
Wild type wings; ebony body 156
Dumpy wings; wild type body 149
Dumpy wings; ebony body 52

You will ultimately explain these results using a Chi-squared test.
(a) Begin by setting up a Punnett Square to predict the phenotypic classes and ratios. Use the results of the Punnett square to form a hypothesis on phenotypic ratios.
(b) Perform the chi-squared analysis. Show work.
(c) What conclusion can be drawn from this analysis?

See the answers below

Explanation:

Assuming that D (d) represents the allele for wing type and E (e) represents the allele for body type, crossing DdEe with DdEe will yield the following offspring according to the Punnet's square (see the attached image):

(a) 9/16 D_E_  wild type

    3/16 D_ee  wild wing, ebony body

    3/16 ddE_ dumpy wing, wild body

    1/16 ddee dumpy wing, ebony body

(b) Chi square = , where O = observed frequency and E = expected frequency.

Phenotype                O                  E                                          

Wild type                 473          9/16 x 830 =  466.875     = 0.08

wild w/ebony b      156          3/16 x 830 = 155.625       =  0.0009

Dumpy w/wild b     149          3/16 x 830 = 155.625       = 0.28

dumpy w/ebony b  52            1/16 x 830 = 51.875         = 0.0003

Total = 0.3612

Degree of freedom = 4 - 1 = 3

Tabulated value of (0.05)= 7.815

(c) Since the calculated Chi square value is less than the tabulated value, we conclude that the observed outcome agrees with the expected outcome and that the cross followed the standard Mendelian pattern.