Hardy-Weinberg Equilibrium and Allele Frequencies

The question asks us to find the frequencies of the two alleles in a diploid population that is in Hardy-Weinberg equilibrium, given the total frequency of homozygotes.

In a population with two alleles, say A and a, the Hardy-Weinberg principle states the relationship between allele frequencies and genotype frequencies. Let the frequency of allele A be denoted by $p$ and the frequency of allele a be denoted by $q$. According to the principle, the sum of allele frequencies in the population is always 1:

$\qquad p + q = 1$

In a population in Hardy-Weinberg equilibrium, the frequencies of the three possible genotypes (AA, Aa, aa) are given by:

• Frequency of homozygous genotype AA = $p^2$
• Frequency of heterozygous genotype Aa = $2pq$
• Frequency of homozygous genotype aa = $q^2$

The sum of the genotype frequencies is also equal to 1:

$\qquad p^2 + 2pq + q^2 = 1$

The question provides the total frequency of homozygotes in the population, which is 0.68. The homozygotes are the individuals with genotype AA ($p^2$) and individuals with genotype aa ($q^2$). Therefore, the sum of their frequencies is:

$\qquad p^2 + q^2 = 0.68$

We have two equations based on the Hardy-Weinberg principle and the given information:

1. $p + q = 1$
2. $p^2 + q^2 = 0.68$

We can solve these two equations simultaneously to find the values of $p$ and $q$. From the first equation, we can express $q$ in terms of $p$ (or vice versa):

$\qquad q = 1 - p$

Substitute this expression for $q$ into the second equation:

$\qquad p^2 + (1 - p)^2 = 0.68$

Expand the equation:

$\qquad p^2 + (1 - 2p + p^2) = 0.68$

Combine like terms:

$\qquad 2p^2 - 2p + 1 = 0.68$

Rearrange the equation to set it equal to zero (standard quadratic form $ap^2 + bp + c = 0$):

$\qquad 2p^2 - 2p + 1 - 0.68 = 0$

$\qquad 2p^2 - 2p + 0.32 = 0$

We can divide the entire equation by 2 to simplify:

$\qquad p^2 - p + 0.16 = 0$

Now we can solve this quadratic equation for $p$. We can use factoring or the quadratic formula. Let's look for two numbers that multiply to 0.16 and add up to 1. The numbers 0.2 and 0.8 fit this requirement ($0.2 \times 0.8 = 0.16$ and $0.2 + 0.8 = 1$). So the equation can be factored as:

$\qquad (p - 0.2)(p - 0.8) = 0$

This gives us two possible solutions for $p$:

$\qquad p - 0.2 = 0 \implies p = 0.2$

$\qquad p - 0.8 = 0 \implies p = 0.8$

If $p = 0.2$, then using $q = 1 - p$, we get $q = 1 - 0.2 = 0.8$. The allele frequencies are 0.2 and 0.8.

If $p = 0.8$, then using $q = 1 - p$, we get $q = 1 - 0.8 = 0.2$. The allele frequencies are 0.8 and 0.2.

Both pairs of values (0.2, 0.8) and (0.8, 0.2) represent the same set of allele frequencies. Let's verify if these frequencies give a total homozygote frequency of 0.68.

If $p = 0.2$ and $q = 0.8$:

$\qquad p^2 + q^2 = (0.2)^2 + (0.8)^2 = 0.04 + 0.64 = 0.68$

This matches the given information. The frequencies of the two alleles are 0.2 and 0.8.

The final answer is the pair of allele frequencies that matches one of the given options.

The frequencies of the two alleles are 0.2 and 0.8.