Question

Suppose that a geneticist discovers & new mutation in Drosophila melanogaster that causes the flies to shake and quiver: She calls this mutation quiver; qu, and determines that it is due to an autosomal recessive gene. She wants tO determine whether the gene encoding quiver is linked to the recessive gene for vestigial wings, Vg: She crosses a fly homozygous for quiver and vestigial traits with a fly homozygous for the wild-type traits_ and then uses the resulting F1 females in a testcross. She obtains the flies from this testcross. Phenotype Number of flies vg qu 230 vg qu 224 Vg qu 97 vg qu 99 Test the hypothesis that the gencs quiver and vestigial assort independently by calculating the chi-squared, X?, for this hypothesis. Provide the X? to one decimal place. X2 Does the X2 value support the hypothesis that the quiver and vestigial genes assort independently Why or why not? Use the partial table of critical values for X2 calculations to test this hypothesis. No, the X? = value indicates that the observed progeny are significantly different from what would be expected with independent assortment of the two genes. No, the X2 = value indicates that there are tOO many phenotypes for independent assortment Yes, the X2 value indicates that the genes vestigial and quiver assort independently: Yes, the X? = value indicates that the observed and expected number of progeny are equal in number:

Answer 1

The student's question involves conducting a chi-squared analysis to test whether two genes in fruit flies assort independently. The calculated chi-squared value is significantly higher than the critical value, indicating that the genes do not assort independently and suggesting linkage between them.

Step-by-step explanation:

The student's question pertains to genetic linkage and the use of chi-squared (X²) analysis to determine whether two genes assort independently in Drosophila melanogaster, also known as fruit flies.

If the genes for quiver and vestigial traits were to assort independently, the expected phenotypic ratio in the testcross offspring would be 1:1:1:1.

However, the data provided (230 vg qu, 224 Vg Qu, 97 vg Qu, 99 Vg qu) suggests an unequal distribution.

To test the hypothesis of independent assortment, we calculate the chi-squared (X²) value. First, we determine the expected number of offspring for each phenotype assuming independent assortment, which would be the total number of offspring divided by four (given the 1:1:1:1 ratio for independent assortment).

With 650 total offspring, we would expect 162.5 offspring for each of the four phenotypes.

Next, we calculate the chi-squared statistic using the formula: X² = Σ[(observed - expected)² / expected]. Plugging the observed values and the expected value (162.5) for each, we get: X² = ((230 - 162.5)² / 162.5) + ((224 - 162.5)² / 162.5) + ((97 - 162.5)² / 162.5) + ((99 - 162.5)² / 162.5) = 167.8

To assess this X² value, we must compare it against the critical value from a chi-squared distribution with a certain degree of freedom. For a test with 2 traits, the degrees of freedom would be (n-1)(m-1) = (2-1)(2-1) = 1. Using a standard chi-squared table, the critical value at 3.841 (p=0.05) for 1 degree of freedom.

Because the calculated X² is much larger than the critical value, we reject the hypothesis that the genes assort independently, indicating linkage.

Answer 2

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.