Question

Suppose peaches have two alleles for fruit color: orange (O) is dominant to yellow (y). A farmer notices the peaches harvested are a mixture of colors:

PhenotypeObservedOrange peaches (OO)554Yellow peaches (oo)365Orange peaches (Oo)444Total1363
Given the data, is the fruit color gene showing Hardy-Weinberg equilibrium? Report your answer to the nearest thousandths (3 numbers after the decimal) with a yes or no afterwards; i.e. if the calculated answer is 8.274 and not in equilibrium, then report 8.274 no
Compare to the expected chi-square value at the 5% level using the given table:
degree of freedom 13.841degree of freedom 25.991degree of freedom 37.851degree of freedom 49.488degree of freedom 511.070

Answer 1

To determine if the fruit color gene is in Hardy-Weinberg equilibrium, we need to calculate the expected genotype frequencies based on the observed phenotype frequencies. Comparing the observed and expected genotype frequencies, we can calculate the chi-square value. Since the calculated chi-square value is much larger than the critical chi-square value, we can conclude that the fruit color gene is not in Hardy-Weinberg equilibrium.

Step-by-step explanation:

To determine if the fruit color gene is showing Hardy-Weinberg equilibrium, we need to calculate the expected genotype frequencies based on the observed phenotype frequencies.

From the data provided, we have:
Orange peaches (OO): 554
Yellow peaches (oo): 365
Orange peaches (Oo): 444
Total: 1363

The allele frequencies can be calculated as follows:
p = (2 * OO + Oo) / (2 * total) = (2 * 554 + 444) / (2 * 1363) = 1552 / 2726 ≈ 0.569
q = (2 * oo + Oo) / (2 * total) = (2 * 365 + 444) / (2 * 1363) = 1174 / 2726 ≈ 0.431

Using the calculated allele frequencies, we can determine the expected genotype frequencies using the Hardy-Weinberg equation:
p^2 + 2pq + q^2 = 1
Expected OO frequency = p^2 = (0.569)^2 ≈ 0.324
Expected oo frequency = q^2 = (0.431)^2 ≈ 0.186
Expected Oo frequency = 2pq = 2 * 0.569 * 0.431 ≈ 0.491

Comparing the observed and expected genotype frequencies, we can calculate the chi-square value using the formula:
chi-square = Σ (observed - expected)^2 / expected

For each genotype, we calculate (observed - expected)^2 / expected:
(554 - 0.324 * 1363)^2 / (0.324 * 1363) ≈ 11286.649
(365 - 0.186 * 1363)^2 / (0.186 * 1363) ≈ 926.334
(444 - 0.491 * 1363)^2 / (0.491 * 1363) ≈ 1461.613

Summing up the calculated values:
chi-square = 11286.649 + 926.334 + 1461.613 ≈ 13674.596

Checking the chi-square table for the degree of freedom (df = number of genotypes - 1), which is 2 - 1 = 1, we find that the critical chi-square value at the 5% level is 3.841.

Since the calculated chi-square value (13674.596) is much larger than the critical chi-square value (3.841), we can conclude that the fruit color gene is not showing Hardy-Weinberg equilibrium.

Therefore, the answer is 13674.596 no.