Allele and Genotype Frequencies

In population genetics, we often describe the genetic makeup of a population by looking at the frequencies of different alleles and genotypes.

For two alleles, say 'A' and 'a', at a specific gene locus, the frequency of these alleles in a population can be represented by $p$ and $q$ respectively.

• Let the frequency of allele 'A' be $p$.
• Let the frequency of allele 'a' be $q$.

Since these are the only two alleles considered for this locus, their frequencies must add up to 1:

\(p + q = 1\)

Genotype Frequency Calculation

According to the Hardy-Weinberg principle, in a non-evolving population, the frequencies of the genotypes can be predicted from the allele frequencies:

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

The sum of these genotype frequencies should also equal 1:

\(p^2 + 2pq + q^2 = (p+q)^2 = 1^2 = 1\)

Maximizing Heterozygote Frequency

The question asks when the frequency of heterozygotes (genotype Aa), which is given by $2pq$, will be highest.

We need to find the values of $p$ and $q$ (where $p+q=1$, and $p, q \ge 0$) that maximize the value of the expression $2pq$.

We can substitute $q = 1 - p$ into the expression for heterozygote frequency:

\text{Frequency of Aa} = 2p(1-p) = 2p - 2p^2

Let's analyze the function $f(p) = 2p - 2p^2$ for values of $p$ between 0 and 1. This is a quadratic function, representing a downward-opening parabola. The maximum value of such a parabola occurs at its vertex.

The $p$-coordinate of the vertex for a quadratic function $ap^2 + bp + c$ is given by $\frac{-b}{2a}$. In our case, $a = -2$ and $b = 2$.

\(p = \frac{-2}{2(-2)} = \frac{-2}{-4} = \frac{1}{2}\)

So, the maximum frequency of heterozygotes occurs when $p = \frac{1}{2}$ (or 0.5).

Since $p + q = 1$, when $p = 0.5$, then $q = 1 - 0.5 = 0.5$.

Thus, the frequency of heterozygotes is highest when the frequency of allele 'A' is equal to the frequency of allele 'a', i.e., $p = q = 0.5$.

At this point, the maximum heterozygote frequency is $2 \times 0.5 \times 0.5 = 2 \times 0.25 = 0.5$.

Checking the Options

Let's consider the given options in light of our finding that heterozygote frequency is maximum when $p=q=0.5$:

• Option 1: Frequency of 'A' is more than frequency of 'a'. This means $p > q$. For example, if $p=0.7, q=0.3$, heterozygote frequency is $2(0.7)(0.3) = 0.42$.
• Option 2: Frequency of 'A' is less than frequency of 'a'. This means $p < q$. For example, if $p=0.3, q=0.7$, heterozygote frequency is $2(0.3)(0.7) = 0.42$.
• Option 3: Frequency of 'A' is equal to frequency of 'a'. This means $p = q$. As we found, this happens when $p=q=0.5$, giving the maximum heterozygote frequency of 0.5.
• Option 4: Frequency of 'A' and 'a' affects the frequency of homozygotes not heterozygotes. This is incorrect. Allele frequencies ($p$ and $q$) directly determine the frequencies of all genotypes, including both homozygotes ($p^2$ and $q^2$) and heterozygotes ($2pq$).

The comparison shows that the frequency of heterozygotes is indeed highest when the frequencies of the two alleles are equal.