In a random mating population of 1000 individuals, 360 belong to genotype $AA$, 480 to genotype $Aa$, and the remaining 160 to genotype $aa$. Calculate the allele frequencies of allele $A$ and allele $a$. Does this population satisfy the Hardy-Weinberg equilibrium? Show step-by-step calculations.

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Accepted Answer

Step 1: Identify given population numbers Total population size ($N$) = 1000 Number of $AA$ individuals = 360 Number of $Aa$ individuals = 480 Number of $aa$ individuals = 160 Step 2: Calculate observed genotype frequencies Observed frequency of $AA$ ($p^2$) = $\frac{360}{1000} = 0.36$ Observed frequency of $Aa$ ($2pq$) = $\frac{480}{1000} = 0.48$ Observed frequency of $aa$ ($q^2$) = $\frac{160}{1000} = 0.16$ Step 3: Calculate allele frequencies ($p$ and $q$) Frequency of allele $A$ ($p$) = $\sqrt{p^2} = \sqrt{0.36} = 0.6$ Frequency of allele $a$ ($q$) = $\sqrt{q^2} = \sqrt{0.16} = 0.4$ Alternative method for allele frequency calculation: Total alleles = $2 	imes 1000 = 2000$ $p = \frac{2(360) + 480}{2000} = \frac{720 + 480}{2000} = \frac{1200}{2000} = 0.6$ $q = \frac{2(160) + 480}{2000} = \frac{320 + 480}{2000} = \frac{800}{2000} = 0.4$ Step 4: Check Hardy-Weinberg Equilibrium $p + q = 0.6 + 0.4 = 1.0$ Expected $p^2 = (0.6)^2 = 0.36$ Expected $2pq = 2 	imes 0.6 	imes 0.4 = 0.48$ Expected $q^2 = (0.4)^2 = 0.16$ 
Since observed genotype frequencies ($0.36, 0.48, 0.16$) exactly match the expected frequencies derived from allele frequencies, the population is in Hardy-Weinberg equilibrium.