Answer :
Okay, let's work through this step-by-step:
Given information:
Total population size: 50 mice
30 black (dominant) mice
20 white (recessive) mice
Step 1: Calculate the allelic frequencies.
The alleles for coat color in this case are B (black) and b (white).
Frequency of B allele = (2 × Number of BB) + (Number of Bb) / (2 × Total population size)
Frequency of B allele = (2 × 15) + 15 / (2 × 50) = 0.6
Frequency of b allele = (2 × Number of bb) + (Number of Bb) / (2 × Total population size)
Frequency of b allele = (2 × 20) + 15 / (2 × 50) = 0.4
Step 2: Calculate the genotypic frequencies.
BB = 15 / 50 = 0.3
Bb = 15 / 50 = 0.3
bb = 20 / 50 = 0.4
Step 3: Validate the results using the chi-square test.
Null hypothesis (H0): The observed genotypic frequencies are consistent with the Hardy-Weinberg equilibrium.
Alternative hypothesis (H1): The observed genotypic frequencies are not consistent with the Hardy-Weinberg equilibrium.
Expected frequencies:
p^2 (BB) = (0.6)^2 = 0.36
2pq (Bb) = 2 × 0.6 × 0.4 = 0.48
q^2 (bb) = (0.4)^2 = 0.16
Chi-square statistic:
χ^2 = Σ [(Observed - Expected)^2 / Expected]
χ^2 = [(15 - 18)^2 / 18] + [(15 - 24)^2 / 24] + [(20 - 8)^2 / 8]
χ^2 = 0.5 + 4.5 + 9
χ^2 = 14
Degrees of freedom = 3 - 1 = 2
At a significance level of 0.05, the critical value of χ^2 with 2 degrees of freedom is 5.991.
Since the calculated χ^2 value (14) is greater than the critical value (5.991), we reject the null hypothesis. The observed genotypic frequencies are not consistent with the Hardy-Weinberg equilibrium.
In summary:
Allelic frequencies:
B = 0.6
b = 0.4
Genotypic frequencies:
BB = 0.3
Bb = 0.3
bb = 0.4
The chi-square test indicates that the observed genotypic frequencies are not consistent with the Hardy-Weinberg equilibrium.
