THE HARDY WEINBERG PRINCIPLE
Used to calculate the frequency of alleles of a gene in a population.
Up till now, we have considered cases where we can predict the frequencies of the different genotypes and phenotypes, e.g. 3A : 1a etc. This is because we have been dealing with the progeny of crosses where we know the genotype of both parents.
When dealing with genes in a population, we cannot predict what genotypes should occur. A couple of examples will show this. Consider the case of the recessive allele for ALBINISM. Albinos are uncommon in the population, so it follows that this particular recessive gene is rare. Similarly, consider the dominant allele which causes BRACHYDACTYLY (deformed digits). In this case, it is the dominant gene which is rare, and the normal recessive gene which is common. While these are extreme examples, they indicate that at a particular locus, any allele may be rare or common. Its frequency is not predictable from knowing whether it is dominant or recessive.
We introduce a new term, the ALLELE FREQUENCY, to take account of this concept. This frequency is an unknown value between 0 and 1. It is frequently given the designation p or q. Where we have a locus with two alleles, say A and a, the usual notation is:
Allele: A a
Frequency: p + q =1
In this case, the allele frequencies are shown as summing to 1, since there are no other alleles. Note that any set of frequencies is applicable to just one population.
The frequency of an allele is not known in advance. Nor is it expected to be constant from one population to another. For example, genes for skin colour, height and facial characteristics show different frequencies in different human populations.
Alleles do not exist in isolation in populations. In most higher organisms individuals are diploid. Just as we can define gene frequencies, so we can define GENOTYPE FREQUENCIES. Just like gene frequencies, these genotype frequencies are unpredictable from locus to locus and from population to population.
While we cannot predict the frequency of a genotype, it turns out that if we know the allele frequency, we can make a reasonable prediction of the genotype frequency. This is based on what is known as the Hardy-Weinberg Law.
If the ALLELE FREQUENCIES at a locus are
Allele: A a
Frequency: p + q = 1
Then the GENOTYPE FREQUENCIES will be:
Genotype: AA Aa aa
Frequency: p2 + 2pq + q2 = 1
A brief derivation of this law can be given as below. We assume that the genes contributed by the female parent (in the ovum) and by the male (in the sperm) are independent. Then:
Since the heterozygote can be produced in two different ways, A from female parent and a from male, or vice versa, a from female parent and A from male, the total frequency of this genotype is p.q + q.p = 2pq. From the table, the three genotypes AA, Aa and aa are thus present in frequencies p2, 2pq and q2.
The above derivation of the Hardy-Weinberg law assumes that the female and male gametes combine at random to produce the genotype. This is similar to assuming the females and males mate at random.
We can use the 2 x 2 checkerboard or the above formula to calculate the expected genotype frequencies i.e. calculate the frequencies of AA (p,2), Aa (2pq) and aa(q2) for example:
If the frequency of A (p) is known to be 0.2 and the frequency of a (q) is 0.8 then:
Frequency of AA (p2) = (0.2)2 = 0.04
Frequency of Aa (2pq) = 2 x 0.2 x 0.8 = 0.32
Frequency of aa (q2) = (0.8)2 = 0.64
If we could recognise all genotypes in a population, then estimating gene frequencies would be easy. We could simply count up the number of each type of gene in the population.
More commonly, if one gene is dominant and one is recessive, we cannot directly count up the number of genes. For example if A is dominant to a, then the above example becomes:
Phenotype: A a
Frequency: p2 + 2pq + q2 = 1
i.e., AA and Aa are both phenotypically A.
We must now estimate the frequency of the gene from the available classes. If the frequency of the recessive class is known, then we can estimate the gene frequency by taking the square root of this.
|EXAMPLE:|| Tongue rollers
TT + Tt
| non- tongue rollers|
|Expected frequency:||p2 + 2pq||q2|
Then the estimated frequency of the gene t, (q) comes from q2 = 0.3
So that q = √0.3 = 0.55
and p = 1-q = 0.45
If there are deviations from random mating in the population sampled, the estimation of gene frequencies is not so straight forward.
- The population is large enough that chance is unlikely to alter the frequencies of the alleles.
- The population is made up of sexually reproducing individuals.
- All individuals are of equal fitness in terms of survivability and reproduction.
- Mating is random with respect to genotype.
- Mutation either does not occur or is in equilibrium.
- Immigration and emigration do not occur.
The consequences of abiding by the Hardy-Weinberg Law are that allele frequencies remain constant from generation to generation.
- Genetic drift
- Natural selection
Non-random mating changes genotype frequencies, but not allele frequencies.
According to the Hardy Weinberg Law the genotype frequencies of
AA = p2
Aa = 2pq
aa = q2
1. In a population in Hardy-Weinberg equilibrium the frequency of albino individuals is 1 in 19,600. Albinism is an autosomal recessive trait, coded for by the allele a. What is the frequency of carriers (heterozygotes) in the population?
Let's calculate the expected frequency of the alleles in the population.
frequency of recessive phenotype (aa) = q2 = 1/19600
frequency of recessive allele (a) = q = 1/140
frequency of dominant allele (A) = p = 1 - (1/140) = 139/140
Now calculate the expected frequency of carriers of this allele.
frequency of carriers (Aa and aA)
= 2 x (139/140) x (1/140)
2. In a hypothetical population of 10,000 which is in Hardy-Weinberg equilibrium 64 000 individuals were found to have the disease myotonic distrophy. Myotonic distrophy is an autosomal dominant trait coded for by the allele M.
Let's calculate the frequency of the recessive allele m in this population.
frequency of dominant phenotype
(MM, Mm and mM) = 64 000/100 000
p2 + 2pq = 64 000/100 000
q2 = 1 - (64 000/100 000)
frequency of allele m q = 0.6
frequency of allele M p = 1 - 0.6 = 0.4
Now calculate the frequency of heterozygotes in this population.
frequency of heterozygotes (Hh and hH) = 2pq = 2 x 0.4 x 0.6 = 0.48