How To Tell If A Population Is In Hardy-weinberg Equilibrium

Understanding whether a population is in Hardy-Weinberg equilibrium (HWE) is a fundamental skill in population genetics. To determine if a population fits this theoretical model, you must compare observed genotype frequencies with those expected under the equilibrium conditions. Which means it serves as the null hypothesis for evolutionary biology, providing a baseline to measure if evolutionary forces—such as natural selection, genetic drift, mutation, gene flow, or non-random mating—are actively shaping allele frequencies. This process involves specific mathematical calculations and statistical testing, which we will explore in detail.

The Theoretical Foundation: Assumptions and Equations

Before diving into the mechanics of testing, it is crucial to understand the five strict assumptions required for a population to remain in equilibrium. If any of these are violated, the population is evolving. The assumptions are:

1. No Mutation: New alleles are not entering the population.
2. Random Mating: Individuals pair by chance, not by genotype or phenotype.
3. No Natural Selection: All genotypes have equal reproductive success.
4. Extremely Large Population Size: Genetic drift is negligible.
5. No Gene Flow: No migration of individuals or gametes into or out of the population.

When these conditions are met, allele frequencies ($p$ and $q$) remain constant across generations, and genotype frequencies stabilize after a single generation of random mating according to the Hardy-Weinberg equation:

$p^2 + 2pq + q^2 = 1$

Where:

• $p$ = frequency of the dominant allele. That said, * $q$ = frequency of the recessive allele. * $p^2$ = expected frequency of homozygous dominant genotype.
• $2pq$ = expected frequency of heterozygous genotype.
• $q^2$ = expected frequency of homozygous recessive genotype.

Additionally, allele frequencies must sum to 1 ($p + q = 1$) Nothing fancy..

Step-by-Step Guide to Testing for Equilibrium

Testing for HWE is a systematic process. You move from raw data (genotype counts) to allele frequencies, then to expected genotype counts, and finally to a statistical verdict It's one of those things that adds up..

Step 1: Collect Observed Genotype Counts

You need a sample of individuals from the population genotyped for a specific locus. Record the number of individuals for each genotype. For a diallelic locus (alleles $A$ and $a$), you will have three counts:

• $N_{AA}$ (Homozygous dominant)
• $N_{Aa}$ (Heterozygous)
• $N_{aa}$ (Homozygous recessive)

Let $N$ be the total number of individuals sampled ($N = N_{AA} + N_{Aa} + N_{aa}$) Surprisingly effective..

Step 2: Calculate Observed Allele Frequencies

Allele frequencies are derived directly from the genotype counts. Because every individual carries two alleles, the total number of alleles in the sample is $2N$ Worth knowing..

• Frequency of allele $A$ ($p$) = $\frac{2(N_{AA}) + N_{Aa}}{2N}$
• Frequency of allele $a$ ($q$) = $\frac{2(N_{aa}) + N_{Aa}}{2N}$

Verify that $p + q = 1$.

Step 3: Calculate Expected Genotype Frequencies

Using the allele frequencies calculated in Step 2 ($p$ and $q$), apply the Hardy-Weinberg equation to find the expected proportions of each genotype under equilibrium.

• Expected frequency of $AA$ = $p^2$
• Expected frequency of $Aa$ = $2pq$
• Expected frequency of $aa$ = $q^2$

Step 4: Calculate Expected Genotype Counts

Convert the expected frequencies into expected counts (numbers of individuals) by multiplying each frequency by the total sample size $N$.

• Expected count $AA$ ($E_{AA}$) = $p^2 \times N$
• Expected count $Aa$ ($E_{Aa}$) = $2pq \times N$
• Expected count $aa$ ($E_{aa}$) = $q^2 \times N$

Step 5: Perform the Chi-Square ($\chi^2$) Goodness-of-Fit Test

This is the standard statistical method to decide if the differences between your observed counts ($O$) and expected counts ($E$) are due to random sampling error or a significant deviation from equilibrium And that's really what it comes down to..

The formula for the test statistic is:

$\chi^2 = \sum \frac{(O - E)^2}{E}$

Calculate this value for each of the three genotype classes and sum them Took long enough..

Step 6: Determine Degrees of Freedom and Critical Value

Degrees of freedom ($df$) for a diallelic locus HWE test is typically 1 Not complicated — just consistent..

Why 1? You have 3 genotype classes ($k=3$). You estimated 1 parameter from the data (the allele frequency $p$, since $q$ is dependent). The formula is $df = k - 1 - \text{parameters estimated} = 3 - 1 - 1 = 1$.

Compare your calculated $\chi^2$ value to the critical value from a Chi-square distribution table at your chosen significance level (usually $\alpha = 0.In real terms, 05$). That's why * **Critical value at $df=1, \alpha=0. Think about it: 05$ is 3. 841.

Step 7: Interpret the Results

• If $\chi^2_{calculated} < 3.841$: Fail to reject the null hypothesis. The population is in Hardy-Weinberg equilibrium (observed deviations are likely due to chance).
• If $\chi^2_{calculated} \geq 3.841$: Reject the null hypothesis. The population is not in Hardy-Weinberg equilibrium. Evolutionary forces are likely acting on this locus.

A Worked Example

Let’s apply these steps to a hypothetical population of 500 plants genotyped for flower color (Red $R$, White $r$).

Observed Counts:

• $RR$ (Red): 245
• $Rr$ (Pink): 210
• $rr$ (White): 45
• Total $N = 500$

1. Allele Frequencies:

• $p (R) = \frac{2(245) + 210}{1000} = \frac{700}{1000} = 0.7$
• $q (r) = \frac{2(45) + 210}{1000} = \frac{300}{1000} = 0.3$
• Check: $0.7 + 0.3 = 1$. ✓

2. Expected Frequencies:

• $RR: p^2 = 0.49$
• $Rr: 2pq = 2(0.7)(0.3) = 0.42$
• $rr: q^2 = 0.09$

3. Expected Counts:

• $E_{RR} = 0.49 \times 500 = 245$
• $E_{Rr} = 0.42 \times 500 = 210$
• $E_{rr} = 0.09 \times 500 = 45$

4. Chi-Square Calculation:

| Genotype | Observed (O) | Expected (E) | $(