Hardy-Weinberg Equilibrium
Introduced| English | Meaning |
|---|---|
| Hardy-Weinberg equilibrium/ˈhɑːdi ˈwaɪnbɜːɡ ˌiːkwɪˈlɪbrɪəm/ | Hardy-Weinberg equilibrium |
What if nothing is changing?
- Evolution is a change in allele frequency — but what if the frequencies hold still?
- Imagine a population where no allele gets more or less common.
- Such a population is not evolving.
- This imaginary steady state is called Hardy-Weinberg equilibrium.
A non-evolving baseline
- Hardy-Weinberg equilibrium 哈迪-温伯格平衡 describes a population whose allele frequencies stay constant.
- It is a theoretical baseline — a picture of no evolution.
- We compare real populations against it.
- If reality does not match, something is causing evolution.
A population in Hardy-Weinberg equilibrium is one that is…
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In Hardy-Weinberg equilibrium, allele frequencies do not change — the population is not evolving.
The equation
- For two alleles, let $p$ be one allele's frequency and $q$ the other's, with $p + q = 1$.
- The genotype frequencies follow $p^2 + 2pq + q^2 = 1$.
- Here $p^2$ is homozygous dominant, $q^2$ homozygous recessive, and $2pq$ heterozygous.
- From an observed $q^2$ you can find $q$, then $p$, then every genotype frequency.
Conditions for no evolution
Step through the conditions a population must meet to stay in Hardy-Weinberg equilibrium - break any one and it evolves.
The Hardy-Weinberg equation is $p^2 + 2pq + q^2 = 1$. What does $q^2$ represent?
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$q^2$ is the frequency of the homozygous recessive genotype (aa); $p^2$ is homozygous dominant, $2pq$ heterozygous.
If $q^2 = 0.09$, what is the recessive allele frequency $q$?
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$q = \sqrt{0.09} = 0.3$. Then $p = 1 - q = 0.7$.
The five conditions
- Equilibrium holds only if no evolutionary force is acting.
- That means: no natural selection, no migration, and no new mutations.
- It also needs random mating and a very large population.
- Break any one of these, and the frequencies shift — the population evolves.
A population under natural selection stays in Hardy-Weinberg equilibrium.
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False. Hardy-Weinberg equilibrium requires no selection; selection changes allele frequencies, so the population evolves and leaves equilibrium.
Select all conditions required for Hardy-Weinberg equilibrium.
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Equilibrium requires no selection, so strong selection would break it. The other three are required.
Hardy-Weinberg is an idealized baseline, not a description of real life. Almost no real population meets all five conditions perfectly. Its value is as a null model: differences from it reveal that evolution is happening.
Finding a hidden allele frequency:
- Suppose $9\%$ of a population shows a recessive trait, so $q^2 = 0.09$.
- Then $q = \sqrt{0.09} = 0.3$, and $p = 1 - 0.3 = 0.7$.
- Now you can predict the carriers: $2pq = 2(0.7)(0.3) = 0.42$, or $42\%$.
Hardy-Weinberg equilibrium is a non-evolving baseline where allele frequencies stay constant, described by $p^2 + 2pq + q^2 = 1$. It holds only under five conditions: no selection, no migration, no mutation, random mating, and a large population. Real populations differ from it — and those differences reveal evolution.
Write from a heterozygote frequency of 0.18
- A population is in Hardy-Weinberg equilibrium.
- The recessive phenotype frequency q squared is 0.01. The lesson case q squared equals 0.09 gives q equals 0.3.
What is the heterozygote frequency 2pq?
q is 0.1 and p is 0.9, so heterozygotes are 0.18.
What is 2pq when q squared is 0.01?
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p and q are 0.9 and 0.1. Other options — 0.3 (the other lesson's q); 0.01 (the recessive phenotype).