How Do You Calculate Genetic Diversity Using Hardy–Weinberg?
Learn how to use Hardy-Weinberg equations to estimate allele and genotype frequencies in a population, and why these do not measure genetic diversity alone.
Use Hardy–Weinberg to estimate allele and genotype frequencies, not to calculate genetic diversity as a single number. In AQA A Level Biology, genetic diversity means the number of different alleles in a population. Hardy–Weinberg tells you how common two alleles and their genotypes are, provided its assumptions apply.
The equations
For a gene with two alleles, let ppp be the frequency of one allele and qqq the frequency of the other:
p+q=1p+q=1p+q=1The predicted genotype frequencies are:
p2+2pq+q2=1p^2+2pq+q^2=1p2+2pq+q2=1Here, p2p^2p2 and q2q^2q2 are the frequencies of the two homozygous genotypes (two copies of the same allele). 2pq2pq2pq is the frequency of the heterozygous genotype (one copy of each allele). A frequency is a proportion of the population, so it lies between 000 and 111.
Worked example
Suppose a recessive phenotype occurs in 16%16\%16% of a population. Assuming Hardy–Weinberg equilibrium and that this phenotype is shown only by people with two recessive alleles:
q2=0.16,q=0.16=0.4q^2=0.16,\qquad q=\sqrt{0.16}=0.4q2=0.16,q=0.16=0.4Then p=1−0.4=0.6p=1-0.4=0.6p=1−0.4=0.6, and the predicted heterozygote frequency is:
2pq=2(0.6)(0.4)=0.482pq=2(0.6)(0.4)=0.482pq=2(0.6)(0.4)=0.48So the estimated allele frequencies are 0.60.60.6 and 0.40.40.4, and 48%48\%48% of the population is predicted to be heterozygous. You have estimated frequencies, not found a numerical value for overall genetic diversity.
What the result means
A higher heterozygote frequency can suggest more variation at one gene because more individuals carry two different alleles. However, it is not the AQA definition of genetic diversity and does not show the number of alleles across the population. To compare populations, scientists may examine allele frequencies at many genes or compare DNA base sequences. In an exam, report each frequency, convert it to a percentage only if required, and state that Hardy–Weinberg is a model based on assumptions.
The common exam mistake is treating 16%16\%16% as qqq. A recessive phenotype gives you q2q^2q2, so take its square root first. Also, do not use the equations as a reliable prediction without considering the model’s assumptions, including a large population, random mating, and no selection, mutation or migration.