In corn, purple kernels (D) are dominant over yellow (d) and smooth kernels (G) are dominant over shrunken (g). and ear of corn has 381 kernels. The data for a dihybrid cross are below:
216 purple, smooth
79 purple, shrunken
65 yellow, smooth
21 yellow, shrunken
Use the chi square formula to test the hypothesis that the two traits segregate and assort independently from generation to generation.
To test the hypothesis of independent assortment, we need to calculate the expected frequencies of each phenotype assuming that the two traits are independently inherited.
First, we need to determine the genotype of each parent. Since purple kernels are dominant over yellow, we know that the purple kernels could be either DD or Dd, while the yellow kernels must be dd. Similarly, smooth kernels could be either GG or Gg, while shrunken kernels must be gg.
Let's use the following symbols to represent each genotype:
- DD: P (for purple)
- Dd: p (for purple)
- dd: q (for yellow)
- GG: S (for smooth)
- Gg: s (for smooth)
- gg: t (for shrunken)
We can then use the product rule to calculate the expected frequencies of each phenotype. For example, the expected frequency of purple and smooth kernels can be calculated as:
P(S) = (P x S) + (P x s) + (p x S) + (p x s)
= (2/3 x 3/4) + (2/3 x 1/4) + (1/3 x 3/4) + (1/3 x 1/4)
= 5/9
Similarly, we can calculate the expected frequencies of the other phenotypes:
- Purple and shrunken: P(t) = (P x t) + (p x t) = (2/3 x 1/4) + (1/3 x 1/4) = 1/6
- Yellow and smooth: q(S) = (q x S) + (q x s) = (1/3 x 3/4) + (1/3 x 1/4) = 1/2
- Yellow and shrunken: q(t) = (q x t) = (1/3 x 1/4) = 1/12
Next, we can use the chi square formula to calculate the chi square statistic:
χ2 = Σ (Observed – Expected)2 / Expected
We have four categories, so the degrees of freedom are:
df = number of categories – 1 = 4 – 1 = 3
Using the observed and expected frequencies, we get:
χ2 = [(216 – 5/9 x 381)2 / (5/9 x 381)] + [(79 – 1/6 x 381)2 / (1/6 x 381)] + [(65 – 1/2 x 381)2 / (1/2 x 381)] + [(21 – 1/12 x 381)2 / (1/12 x 381)]
= 109.70
Since the chi square value is higher than the critical value at a significance level of 0.05 for 3 degrees of freedom (7.815), we reject the null hypothesis of independent assortment. This means that the two traits are not assorting independently and are likely linked on the same chromosome.