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Exp 1: Testing Hardy-Weinberg Equilibrium & Chi-Square (chi^2) Goodness-of-Fit in Natural and Model Populations

Experiment 01Population GeneticsStatistical Testing

Testing Hardy-Weinberg Equilibrium & Chi-Square ($chi^2$) Goodness-of-Fit in Natural and Model Populations

Duration: 4 Hours • Wet Lab / Model Simulation
๐ŸŽฏ Experimental Aim & Learning Objectives
To determine allele and genotypic frequencies for a diallelic locus (e.g., human PTC taste sensitivity or model population beads); test whether the observed genotype frequencies conform to the Hardy-Weinberg expectations ($p^2 + 2pq + q^2 = 1$); and evaluate deviations using Pearson’s Chi-Square ($chi^2$) goodness-of-fit test at $text{df} = 1$.
๐Ÿ”ฌ Theoretical Foundation & Mathematical Principles

The Hardy-Weinberg Principle forms the null model of evolutionary biology. In an infinitely large, panmictic (randomly mating) diploid population free of mutation, migration, natural selection, and genetic drift, allele and genotype frequencies remain constant across generations:

  • Binomial Expansion: For a locus with two alleles $A$ (frequency $p$) and $a$ (frequency $q$), where $p + q = 1$:
    $$(p + q)^2 = p^2 + 2pq + q^2 = 1$$
    where $p^2 = text{Freq}(AA)$, $2pq = text{Freq}(Aa)$, and $q^2 = text{Freq}(aa)$.
  • Allele Frequency Calculation (Direct Gene Counting):
    $$p = frac{2 cdot N_{AA} + N_{Aa}}{2N}, quad q = frac{2 cdot N_{aa} + N_{Aa}}{2N}$$
  • Pearson’s Chi-Square ($chi^2$) Goodness-of-Fit:
    $$chi^2 = sum frac{(O – E)^2}{E} = frac{(N_{AA} – N p^2)^2}{N p^2} + frac{(N_{Aa} – 2Npq)^2}{2Npq} + frac{(N_{aa} – N q^2)^2}{N q^2}$$
    The degrees of freedom is: $text{df} = k – 1 – m = 3 – 1 – 1 = 1$, where $k=3$ genotypic classes and $m=1$ estimated independent parameter ($p$).
  • Decision Rule: The critical value at $alpha = 0.05$ with $text{df} = 1$ is $chi^2_{text{critical}} = 3.841$. If $chi^2 > 3.841$, the null hypothesis of equilibrium is rejected, demonstrating the action of evolutionary forces or non-random mating.
HARDY-WEINBERG EQUILIBRIUM & CHI-SQUARE (χ²) GOODNESS-OF-FITHardy-Weinberg Genotype Frequencies (p² + 2pq + q² = 1)Allele Frequency p (A) [0.0 → 1.0] • q (a) [1.0 → 0.0]Genotype Frequency →1.00.50.0p² (AA)q² (aa)2pq (Aa) Max = 0.50p = q = 0.5CHI-SQUARE (χ²) TEST OF EQUILIBRIUM1. Pearson’s χ² Test Statistic:χ² = ∑ [ (Observed − Expected)² / Expected ]2. Degrees of Freedom (df) in HWE:df = No. of Genotypic Classes − No. of Alleles = 3 − 2 = 13. Critical Decision Rules at df = 1 (α = 0.05):

• χ²_critical (p = 0.05, df = 1) = 3.841• If χ² ≤ 3.841: Fail to reject H₀ → Population conforms to HWE• If χ² > 3.841: Reject H₀ (p < 0.05) → Evolutionary forces acting(Forces: Mutation, Migration, Selection, Drift, Assortative Mating)

Figure 1.1: Hardy-Weinberg equilibrium parabolic genotype frequency distribution as a function of allele frequencies ($p, q$), and Pearson’s Chi-Square ($chi^2$) goodness-of-fit decision threshold at $text{df} = 1$.
๐Ÿงช Materials & Experimental Assay
  • Phenotyping Assay: Phenylthiocarbamide (PTC) taste test paper strips (impregnated with $5text{ mu g}$ PTC) and control untreated paper strips; sanitized disposable forceps.
  • Model Alternative: Two-color bead simulation kits (400 red beads representing allele $A$, 600 white beads representing allele $a$, total $N = 500$ diploid individuals).
๐Ÿ“‹ Step-by-Step Experimental Protocol
  1. Human Phenotyping (PTC Tasting): Place a control paper strip on the tongue, followed by a PTC paper strip. Score individuals as Tasters (dominant phenotype, $TT$ or $Tt$, experiencing intensely bitter taste) or Non-Tasters (recessive phenotype, $tt$, tasting only cellulose). Record total sample size ($N$).
  2. Model Sampling (Direct Counting): In a model population with codominance (or bead drawing), draw 100 pairs of beads randomly without replacement to form 100 diploid genotypes: $AA$, $Aa$, and $aa$.
  3. Frequency Derivation: Calculate observed counts $N_{AA}$, $N_{Aa}$, and $N_{aa}$. Compute allele frequencies $p$ and $q$ by direct gene counting.
  4. Expected Genotypic Numbers: Compute expected numbers: $E(AA) = N cdot p^2$, $E(Aa) = N cdot 2pq$, and $E(aa) = N cdot q^2$.
  5. Chi-Square Testing: Calculate $chi^2 = sum frac{(O – E)^2}{E}$. Compare with $chi^2_{0.05, 1} = 3.841$ and state biological conclusion.
๐Ÿ“Š Representative Data & Goodness-of-Fit Analysis
Genotype ClassObserved ($O$)Expected ProportionExpected ($E$)$(O – E)$$(O – E)^2 / E$
$AA$ (Homozygous Dominant)140$p^2 = (0.60)^2 = 0.360$144.0-4.00.111
$Aa$ (Heterozygote)200$2pq = 2(0.6)(0.4) = 0.480$192.0+8.00.333
$aa$ (Homozygous Recessive)60$q^2 = (0.40)^2 = 0.160$64.0-4.00.250
Total ($N = 400$)4001.000400.00.0$chi^2 = 0.694$

Statistical Inference: Calculated $chi^2 = 0.694 < chi^2_{0.05, 1} = 3.841$ ($p = 0.405$). We fail to reject the null hypothesis; the population is in Hardy-Weinberg equilibrium at this locus.

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