In a Drosophila dihybrid cross a sepia male was mated to a a

In a Drosophila dihybrid cross, a sepia male was mated to a apterous female. Their phenotypically wild F_1 female produced the following offspring when crossed to phenotypically wild F_1 males: Phonotype Number wild type 450 sepia 160 apterous 140 sepis/apterous 50 Please set up a null hypothesis and use a Chi-square analysis to see if the observation follows the Mendelian phenotypic ratio.

Solution

In dihybrid cross, expected phenotypic ratio is 9:3:3:1. Chi square test is also called goodness of fit. In this test the difference between observed and expected frequency will measured.

The hypothesis

H0 (wild type) = 9/16 = .56. ( No effect if follow 9:3:3:1 ratio)

(Sepia) = 3/16 = .19

(Apterous) = 3/16= .19

(Sepia/apterous) = 1/16 = .062

H1 = if not follow this ratio

Total population N= 800 (450+160+140+50)

Wild type = 450

Expected (E) = .56 *800= 448

Sepia = .19*800= 152

Apterous = .19*800 = 152

Sepia/apterous = .062*800= 49

Chi square = (observed-expected)2/expected

Here,

(450-448)2/448+(160-152)2/152+(140-152)2/152+(50-49)2/49

= .008+.42+.947+.02

= 1.395

Degree of freedom = 4 elements

n-1= 4-1=3

So from table

p (.95) >.35

Null hypothesis is not rejected

This data conclude that proportion of offsprings are not statistically significantly different from their hypothesized value. This data is consistant with phenotypic mandalian ratio9:3:3:1.

If considered p(.5) then the value statistically significant , in this case null hypothesis will be rejected.