Answer:
See the answer below
Explanation:
Using the formula for calculating Chi square ([tex]X^2[/tex]):
[tex]X^2[/tex] = [tex]\frac{(O - E)^2}{E}[/tex] where O = observed frequency and E = expected frequency.
The observed frequencies for the four phenotypes are 88, 62, 62, and 81 respectively.
For the expected frequency, the phenotypes are supposed to assort in 9:3:3:1 according to Mendel ratio.
Hence, expected frequencies are calculated as:
phenotype (1) = 9/16 x 293 = 164.81
phenotype (2) = 3/16 x 293 = 54.94
phenotype (3) = 3/16 x 293 = 54.94
phenotype (4) = 1/16 x 293 = 18.31
The [tex]X^2[/tex] is calculated thus:
Phenotype O E [tex]X^2[/tex]
1 88 164.81 [tex]\frac{(88 - 164.81)^2}{164.81}[/tex] = 35.80
2 62 54.94 [tex]\frac{(62 - 54.94)^2}{54.94}[/tex] = 0.91
3 62 54.94 [tex]\frac{(62 - 54.94)^2}{54.94}[/tex] = 0.91
4 81 18.31 [tex]\frac{(81 - 18.31)^2}{18.31}[/tex] = 214.64
Total [tex]X^2[/tex] = 35.80 + 0.91 + 0.91 + 214.64 = 252.26
To the nearest tenth = 252
Degree of freedom = n - 1 = 4- 1 = 3
[tex]X^2[/tex] tabulated = 7.815
The calculated [tex]X^2[/tex] exceeds the critical value, hence, the null hypothesis is rejected. The two genes did not assort independently.
