Allele frequencies and the Hardy-Weinberg principle

A population and its gene pool

A population is a group of individuals of one species that live in one place and breed with one another. All the genes of all its members, with all their alleles, make up its gene pool. An allele is one of the forms of a gene. A gene for flower colour may have an allele $R$ for red and an allele $r$ for white.

The allele frequency is the share of all the copies of a gene in the gene pool that are that allele. Count copies, not individuals. A diploid individual has two copies of each gene. An $AA$ individual has two copies of $A$. An $Aa$ individual has one copy of each. An $aa$ individual has two copies of $a$.

Take a sample of 100 plants: 36 are $AA$, 48 are $Aa$ and 16 are $aa$.

• Copies of $A$: $2 \times 36 + 48 = 120$ out of $200$ copies. So the frequency of $A$ is $0.6$.
• Copies of $a$: $2 \times 16 + 48 = 80$ out of $200$ copies. So the frequency of $a$ is $0.4$.

We write the frequency of $A$ as $p$ and of $a$ as $q$. For two alleles, $p + q = 1$.

The Hardy-Weinberg principle

In 1908, the English mathematician G. H. Hardy and the German doctor W. Weinberg showed, independently, what happens to allele frequencies when nothing disturbs a population.

NCERT states the Hardy-Weinberg principle like this. The allele frequencies in a population are stable and stay constant from generation to generation. The gene pool stays constant. This is called genetic equilibrium, or Hardy-Weinberg equilibrium. The total of all the allele frequencies is 1.

For one gene with two alleles in a diploid population, the genotype frequencies are given by the expansion of $(p + q)^2$.

Here $p^2$ is the frequency of $AA$, $2pq$ is the frequency of $Aa$, and $q^2$ is the frequency of $aa$.

Where the equation comes from

Think of the gene pool as a big bowl of gametes. A share $p$ of the eggs carry $A$, and a share $q$ carry $a$. The same is true of the sperms. If eggs and sperms meet at random, the chance of each genotype is a product.

• $AA$: $p \times p = p^2$.
• $Aa$: $p \times q$ for an $A$ egg with an $a$ sperm, and $q \times p$ for an $a$ egg with an $A$ sperm. The two add up to $2pq$.
• $aa$: $q \times q = q^2$.

The square has a side of length 1. Its area is 1, and its four parts are the four ways to meet. After one generation of random mating, the genotypes are in these proportions. The allele frequencies have not changed. $p$ is still $p$.

What disturbs the equilibrium

The principle holds only while five factors stay away. NCERT names them.

• Gene migration or gene flow: individuals move in or out and carry alleles with them.
• Genetic drift: chance changes in the allele frequencies, mostly in small populations.
• Mutation: a change in the DNA makes a new allele.
• Genetic recombination: alleles are shuffled into new combinations. NCERT counts this as a factor. Many A-Level books point out that recombination alone does not change the frequency of an allele.
• Natural selection: some genotypes leave more young than others.

A book of A-Level or IB lists the conditions as a population that is very large, mates at random, and has no mutation, no migration and no selection. These come to the same thing.

If the frequencies of the alleles change from generation to generation, the equilibrium is disturbed. NCERT interprets this change as evolution. So the principle is a yardstick: it tells us what to expect when nothing happens, and a departure from it shows that something is happening.

This part is not in the NEET syllabus, but A-Level and IB ask it. To test whether real counts fit the Hardy-Weinberg numbers, the students use a chi-square test.

Using the principle

The hidden carriers. A recessive allele is hidden in the people who carry it but are healthy. Suppose that a recessive condition affects 1 person in 2500.

1. The affected people are $aa$. So $q^2 = \tfrac{1}{2500}$, and $q = \tfrac{1}{50} = 0.02$.
2. Then $p = 1 - 0.02 = 0.98$.
3. The carriers are $Aa$, with the frequency $2pq = 2 \times 0.98 \times 0.02 = 0.0392$.

About $3.9$ people in 100 are carriers, while only $0.04$ in 100 are affected. There are about 98 carriers for each affected person. This is why a harmful recessive allele is so hard to remove from a population.

Why we start from $q^2$. Only the recessive genotype shows its genotype directly. A person with the dominant trait may be $AA$ or $Aa$. So the frequency of the recessive trait is the starting point, and the square root of it is $q$.

You should see that the heterozygotes are the most common genotype when $q$ is near one half, and that when $q$ is small there are many carriers for each affected person.

Try it in Python

The program uses exact fractions to compute the three genotype frequencies for $p = 0.7$.

Practise with new numbers

Common mistakes

• Counting individuals, not gene copies. Each individual has two copies of the gene. Count them.
• Taking $q^2$ for $q$. If 1 person in 100 is affected, $q^2 = 0.01$ and $q = 0.1$.
• Forgetting the factor 2 in $2pq$. The heterozygote can form in two ways.
• Saying that the dominant allele becomes more common because it is dominant. Dominance does not change allele frequencies.
• Thinking that the principle says a population never evolves. It says what happens when none of the five factors acts.

Remember these

• Population, gene pool, allele frequency: $p + q = 1$.
• Hardy-Weinberg: $p^2 + 2pq + q^2 = 1$. The frequencies stay constant from generation to generation when nothing disturbs them.
• The five factors: gene flow, genetic drift, mutation, genetic recombination, natural selection.
• A change in allele frequency is read as evolution.
• Start from $q^2$, the recessive trait, to find $q$, $p$ and the carriers $2pq$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What does $p + q$ equal? What does $p^2 + 2pq + q^2$ equal?

2. In a sample of 50 $AA$, 40 $Aa$ and 10 $aa$ individuals, what are $p$ and $q$?

3. A recessive condition affects 1 baby in 900. Find $q$ and the frequency of carriers as a fraction.

4. Name the five factors that disturb the Hardy-Weinberg equilibrium.

5. A population is in equilibrium and $p = q = 0.5$. What fraction of the individuals is heterozygous?

References

• NCERT (2023). Biology, Textbook for Class XII, chapter 6.
• Clark M. A., Douglas M. and Choi J. (2018). Biology 2e, OpenStax, chapter 19.2, Population Genetics.