Answer:
See the answer below
Explanation:
Null hypothesis: The two genes are independently assorting and therefore, are not linked to each other.
Let A represents the allele for tail length and B for fur color. Long-tail A is dominant over short tail a while brown fur B is dominant over white fur b.
For the first cross involving true-breeding mice:
AABB x aabb
F1 AaBb (long tails and brown fur)
For the second cross:
AaBb x aabb
4 AaBb - Long-tail, brown fur
4 Aabb - Long tail, white fur
4 aaBb - short tail, brown fur
4 aabb - short tail, white fur
Since the phenotypic ratio from the cross is 1:1:1:1, if the null hypothesis was to be true, it means that the expected phenotype ratio should be 1:1:1:1.
In order to test this hypothesis, we use Chi-square:
phenotype O E X^2
Long-tail, brown fur 118 97.5 = 4.31
Long tail, white fur 77 97.5 4.31
short tail, brown fur 81 97.5 2.79
short tail, white fur 114 97.5 2.79
Total 14.2
Degree of freedom = 4 - 1 = 3
Critical Chi-square value = 7.815
The calculated Chi-square value is more than the critical value, hence, the null hypothesis is rejected.
