Question

22. A plant geneticist is investigating the inheritance of genes for bitter taste (Su) and explosive rind (e) in watermelon (Citrullus lanatus). The explosive rind is recessive and causes watermelons to burst when cut. Non-bitter watermelons are a result of the recessive genotype (susu). The geneticist wishes to determine if the genes assort independently. She performs a testcross between a bitter/nonexplosive hybrid and a plant homozygous recessive for both traits. The following offspring are produced:
Bitter/nonexplosive 88
Bitter/explosive 62
Non-bitter/non-explosive 62
Non-bitter/explosive 81

Calculate the chi-square value and also the chi-square of independence for the null hypothesis that the two genes assort independently. Calculate using a chi-square test and the chi-square test of independence. Give your answer to the nearest tenth.

Answer 1

Answer:

See the answer below

Explanation:

Using the formula for calculating Chi square ($X^2$):

$X^2$  =  $\frac{(O - E)^2}{E}$ 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 $X^2$ is calculated thus:

Phenotype            O               E                             $X^2$

  1                          88            164.81            $\frac{(88 - 164.81)^2}{164.81}$ = 35.80

  2                          62             54.94            $\frac{(62 - 54.94)^2}{54.94}$ = 0.91

  3                           62             54.94              $\frac{(62 - 54.94)^2}{54.94}$ = 0.91

  4                            81              18.31               $\frac{(81 - 18.31)^2}{18.31}$ = 214.64

Total $X^2$ = 35.80 + 0.91 + 0.91 + 214.64 = 252.26

To the nearest tenth = 252

Degree of freedom = n - 1 = 4- 1 = 3

$X^2$ tabulated  = 7.815

The calculated $X^2$ exceeds the critical value, hence, the null hypothesis is rejected. The two genes did not assort independently.