- Given allele frequencies \(p = 0.2\) and \(q = 0.8\), the Hardy-Weinberg genotype frequencies are calculated as \(p^2, 2pq, q^2\).
- The frequency of the homozygous dominant genotype \(f(HH) = p^2 = (0.2)^2 = 0.04\).
- The frequency of the heterozygous genotype \(f(Hh) = 2pq = 2 \times 0.2 \times 0.8 = 0.32\).
- The frequency of the homozygous recessive genotype \(f(hh) = q^2 = (0.8)^2 = 0.64\).
- The results are verified as \(0.04 + 0.32 + 0.64 = 1\), with final frequencies \(\boxed{f(HH)=0.04, f(Hh)=0.32, f(hh)=0.64}\).
Explanation
- Analysis of Hardy-Weinberg Equilibrium
In a population following the Hardy-Weinberg equilibrium, the distribution of genotypes is determined by the allele frequencies. We are given the frequency of the dominant allele \(H\) as \(p = 0.2\) and the frequency of the recessive allele \(h\) as \(q = 0.8\). The total allele frequency is \(p + q = 0.2 + 0.8 = 1\), which satisfies the fundamental requirement for the Hardy-Weinberg model. We seek to find the frequencies of the three possible genotypes: homozygous dominant (\(HH\)), heterozygous (\(Hh\)), and homozygous recessive (\(hh\)). These frequencies are given by the terms of the expansion of \((p + q)^2 = p^2 + 2pq + q^2 = 1\).
- Calculating Homozygous Dominant Frequency
To calculate the frequency of the homozygous dominant genotype \(f(HH)\), we use the formula \(f(HH) = p^2\). Substituting the given value of \(p = 0.2\), we have:
\[f(HH) = (0.2)^2 = 0.04\]
This means that 4% of the population is expected to be homozygous dominant for this trait.
- Calculating Heterozygous Frequency
Next, we calculate the frequency of the heterozygous genotype \(f(Hh)\) using the formula \(f(Hh) = 2pq\). Substituting the values \(p = 0.2\) and \(q = 0.8\):
\[f(Hh) = 2 \times 0.2 \times 0.8 = 0.32\]
Thus, 32% of the population is expected to be heterozygous.
- Calculating Homozygous Recessive Frequency and Verification
Finally, we determine the frequency of the homozygous recessive genotype \(f(hh)\) using the formula \(f(hh) = q^2\). Substituting the value \(q = 0.8\):
\[f(hh) = (0.8)^2 = 0.64\]
This indicates that 64% of the population is expected to be homozygous recessive. We verify our results by ensuring the sum of the frequencies equals 1:
\[f(HH) + f(Hh) + f(hh) = 0.04 + 0.32 + 0.64 = 1.00\]
The calculated genotype frequencies are: \(f(HH) = 0.04\), \(f(Hh) = 0.32\), and \(f(hh) = 0.64\).
Examples
The Hardy-Weinberg principle is essential in population genetics for understanding how allele frequencies remain constant in the absence of evolutionary forces. For instance, conservation biologists might use this model to estimate the prevalence of genetic diseases in endangered animal populations. By observing the frequency of individuals expressing a recessive trait, they can infer the proportion of carriers (heterozygotes) in the population, which helps in developing effective strategies for species preservation and monitoring genetic diversity.