The correct answer is : No, the χ² value does not support the hypothesis that the quiver and vestigial genes assort independently. To test whether the genes for quiver (qu) and vestigial wings (Vg) assort independently, you can use a chi-squared (χ²) test. The null hypothesis (H0) is that these genes assort independently, while the alternative hypothesis (Ha) is that they do not assort independently.
First, let's calculate the expected number of offspring for each phenotype if the genes assort independently based on the Punnett square:
1. vg qu: (1/4) * Total number of offspring = (1/4) * (230 + 224 + 97 + 99) = 162.5
2. vg Vg: (1/4) * Total number of offspring = (1/4) * (230 + 224 + 97 + 99) = 162.5
3. Vg qu: (1/4) * Total number of offspring = (1/4) * (230 + 224 + 97 + 99) = 162.5
4. Vg Vg: (1/4) * Total number of offspring = (1/4) * (230 + 224 + 97 + 99) = 162.5
Now, let's calculate the chi-squared (χ²) value using the formula:
χ² = Σ((Observed - Expected)² / Expected)
Using the given data:
χ² = [(230 - 162.5)² / 162.5] + [(224 - 162.5)² / 162.5] + [(97 - 162.5)² / 162.5] + [(99 - 162.5)² / 162.5]
χ² = (67.5² / 162.5) + (61.5² / 162.5) + (−65.5² / 162.5) + (−63.5² / 162.5)
Now, calculate the χ² value:
χ² ≈ 101.25 + 92.25 + 100.83 + 97.23 ≈ 391.56 (rounded to two decimal places)
Now, Compare this χ² value to the critical value from the chi-squared distribution table at a specific significance level and degrees of freedom (df). The degrees of freedom for this test are (4 - 1) * (2 - 1) = 3.
Let's assume a significance level of 0.05 (which is commonly used). Use a chi-squared distribution table or use a calculator to find the critical value for χ² with 3 degrees of freedom at a significance level of 0.05.
Assuming that the critical value at a significance level of 0.05 is approximately 7.815 (you can verify this from a chi-squared distribution table):
Since the calculated χ² value (391.56) is much greater than the critical value (7.815), we reject the null hypothesis (H0). Therefore, we conclude that the genes for quiver and vestigial wings do not assort independently.
So, the correct answer is: No, the χ² value does not support the hypothesis that the quiver and vestigial genes assort independently.