Prompt Title: Pea Gene Linkage Analysis.

To determine if the genes are linked, we need to calculate the expected number of offspring for each possible genotype if the genes are not linked, and compare those expected values to the observed values using a chi-square test. The null hypothesis is that the genes are not linked, while the alternative hypothesis is that they are linked.

First, we need to determine the genotypes of the parents and the F1 offspring. The orange-podded parent with virus sensitivity must be homozygous recessive for both traits (ppmo/mo), while the green-podded parent with virus resistance must be homozygous dominant for both traits (PPMo/Mo). The F1 offspring will therefore all be heterozygous for both traits (PpMo/mo).

Using the Punnett square, we can predict the expected phenotypic ratios of the offspring from the cross between the orange-podded parent and the green-podded parent:

The expected phenotypic ratios for the testcross between the F1 and the orange-podded virus-sensitive parent are:

Using these ratios, we can calculate the expected number of offspring for each genotype:

To calculate the chi-square value, we can use the formula:

χ² = Σ((observed - expected)² / expected)

where Σ represents the sum of the values for each category.

Using the observed and expected values for the testcross, we get:

χ² = [(145-87.5)²/87.5] + [(160-87.5)²/87.5] + [(46-87.5)²/87.5] + [(42-87.5)²/87.5] χ² = 88.69

The degrees of freedom for this test are (2-1) x (2-1) = 1, since there is only one linkage group. Using a chi-square table or calculator with one degree of freedom and a significance level of 0.05, we find that the critical value is 3.84.

Since the calculated chi-square value (88.69) is greater than the critical value (3.84), we reject the null hypothesis and conclude that the genes are linked.