In the hypothesis the phenotypic differences are due to two alleles of one gene. The expected results are 3:1 ratio in the F2 progeny otherwise 300 purple and 100 blue.
So, we should use the formula, x2 = ∑ (observed – expected)2 / expected.
Table:
| Class | Observed(O) | Expected(E) | (O-E) 2 | (O-E) 2 /E |
| Purple | 320 | 300 | 400 | 1.33 |
| Blue | 80 | 100 | 400 | 4.00 |
The x2 = 5.33
The table gives the probability, whether the data fits the stated hypothesis.
We have to determine the degrees of freedom or the number of independent variables in the data.
The degrees of freedom = (number phenotype in the problem – 1).
In this problem, the number of phenotypes is two (purple and blue), so the degree of freedom is 1.
Use the table and find the p value, for the data. It is between 0.025 and 0.01. The p value is the probability that the deviation of the observed from the expected. It is due to only chance.
The relative standard used commonly in research in biology, for the rejection of hypothesis is p<0.05.
Thus, in this case, the data do not support the hypothesis.