A population of 150 individuals has an allele frequency of 0.2 for the dominant allele $(H)$ and a frequency of 0.8 for the recessive allele $(h)$. Use the Hardy-Weinberg equation to determine the frequencies of the three genotypes:

$\begin{array}{l}

f(HH)=p^2=\square \\

f(Hh)=2 pq=\square \\

f(hh)=q^2=\square

\end{array}$

Verified Answer :

  • The allele frequencies are given as \(p = 0.2\) and \(q = 0.8\).
  • The genotype frequencies are calculated using the Hardy-Weinberg equilibrium formula \((p + q)^2 = p^2 + 2pq + q^2 = 1\).
  • The calculations yield: \(f(HH) = p^2 = 0.04\), \(f(Hh) = 2pq = 0.32\), and \(f(hh) = q^2 = 0.64\).
  • Final genotype frequencies: \(\boxed{f(HH)=0.04, f(Hh)=0.32, f(hh)=0.64}\).

Explanation

  1. Understanding the Problem and Data

The problem asks us to determine the genotype frequencies of a population of 150 individuals in Hardy-Weinberg equilibrium, given the allele frequencies for the dominant trait (H) and the recessive trait (h).
The frequency of the dominant allele (H) is \(p = 0.2\), and the frequency of the recessive allele (h) is \(q = 0.8\).
According to the Hardy-Weinberg equilibrium, for a gene with two alleles, the sum of their frequencies must be 1, which is satisfied here: \(p + q = 0.2 + 0.8 = 1.0\).
The genotype frequencies are derived from the binomial expansion \((p + q)^2 = p^2 + 2pq + q^2 = 1\), where \(f(HH) = p^2\), \(f(Hh) = 2pq\), and \(f(hh) = q^2\).

  1. Calculating Genotype Frequencies

Using the given allele frequencies, we calculate the expected frequencies for each genotype:

  • For homozygous dominant individuals, \(f(HH) = p^2 = (0.2)^2 = 0.04\).
  • For heterozygous individuals, \(f(Hh) = 2pq = 2 \times (0.2) \times (0.8) = 0.32\).
  • For homozygous recessive individuals, \(f(hh) = q^2 = (0.8)^2 = 0.64\).

We can verify these calculations by summing the frequencies: \(0.04 + 0.32 + 0.64 = 1.0\), which confirms that the total population distribution sums to 1.

  1. Final Results

Based on the Hardy-Weinberg calculations, the final frequencies for the genotypes are:

  • \(f(HH) = p^2 = 0.04\)
  • \(f(Hh) = 2pq = 0.32\)
  • \(f(hh) = q^2 = 0.64\)

These values represent the proportions of the population of 150 individuals that would be expected to have each respective genotype.

Examples

In genetics, the Hardy-Weinberg principle serves as a null hypothesis to understand evolution. By calculating expected genotype frequencies like \(f(HH)=0.04\), \(f(Hh)=0.32\), and \(f(hh)=0.64\), biologists can determine if a population is evolving. If the observed numbers in a population significantly deviate from these calculated frequencies, it suggests that factors such as natural selection, non-random mating, or migration are occurring within the population. This model is a foundational tool in population genetics, helping scientists analyze inheritance patterns and predict how trait frequencies may change over generations.