Null Hypothesis and Hardy-Weinberg Equilibrium

Understanding the Null Hypothesis and Hardy-Weinberg Equilibrium

In scientific inquiry, a null hypothesis (H_0) serves as a baseline for comparison.
It represents the hypothesis where "nothing is happening," meaning there is no change, no effect, or everything remains constant.
Purpose: It provides a point of comparison to determine if any observed difference is statistically significant or merely due to random chance.
Example:
- Null Hypothesis (H_0): Cats show no preference for food based on shape or texture.
- Alternative Hypothesis (H_1): Cats do show a preference for food based on shape or texture.
- If experimental results significantly deviate from the null hypothesis, we reject the null in favor of the alternative.

Similarly, in evolutionary biology, we have a null model for evolution.
This null model assumes no evolution is occurring within a population.
This specific null model is known as Hardy-Weinberg Equilibrium.
Under HWE, the gene pool remains stable over time, and allele frequencies do not change from one generation to the next.

For a population to be in Hardy-Weinberg Equilibrium, meaning no evolution is occurring, the following five conditions must be true:

No Mutation: No new alleles are created, nor are existing alleles changed.
No Natural Selection: All individuals have equal survival and reproductive rates, meaning no allele provides a selective advantage.
No Gene Flow (Migration): There is no movement of alleles into or out of the population.
Large Population Size: The population must be large enough to prevent random fluctuations in allele frequencies due to chance events (i.e., no genetic drift).
Random Mating: Individuals choose mates without preference for specific genotypes, ensuring that all allele combinations occur in proportion to their frequencies.

To assess if a population is evolving, we compare observed frequencies to what is expected under Hardy-Weinberg Equilibrium.

Allele Frequencies: Represent the proportion of a specific allele in the gene pool.
- Let p be the frequency of the dominant allele (e.g., big A, A).
- Let q be the frequency of the recessive allele (e.g., little a, a).
- The sum of allele frequencies must equal one: p + q = 1
Genotype Frequencies (Expected under HWE): Represent the proportion of each genotype in the population if HWE holds.
- Frequency of homozygous dominant (AA): p^2
- Frequency of heterozygous (Aa): 2pq
- Frequency of homozygous recessive (aa): q^2
- The sum of genotype frequencies must equal one: p^2 + 2pq + q^2 = 1

We use a population of moths to walk through the steps of determining if evolution is occurring.

Given: counts of individuals for each genotype.
- Assume 200 individuals total.
- Each individual has 2 alleles, so there are 2 imes 200 = 400 alleles in the population.
Example Calculation for Allele Frequencies:
- If we have 61 homozygous dominant (AA), 99 heterozygous (Aa), and 40 homozygous recessive (aa) individuals.
- Frequency of allele A (p):
  p = rac{(2 imes ext{Number of AA individuals}) + ( ext{Number of Aa individuals})}{ ext{Total number of alleles}}
  p = rac{(2 imes 61) + 99}{400} = rac{122 + 99}{400} = rac{221}{400} = 0.5525 ext{ (approximately } 0.553)
- Frequency of allele a (q): We can calculate this in two ways:
  1. q = 1 - p = 1 - 0.553 = 0.447
  2. q = rac{(2 imes ext{Number of aa individuals}) + ( ext{Number of Aa individuals})}{ ext{Total number of alleles}}
    q = rac{(2 imes 40) + 99}{400} = rac{80 + 99}{400} = rac{179}{400} = 0.4475 ext{ (approximately } 0.447)}

Given: New counts of individuals for each genotype in Generation 2.
- Assume total individuals = 200.
Example Calculation: If in Generation 2, there are 60 AA, 100 Aa, and 40 aa individuals.
- Observed frequency of AA: rac{60}{200} = 0.30
- Observed frequency of Aa: rac{100}{200} = 0.50
- Observed frequency of aa: rac{40}{200} = 0.20

Use the allele frequencies calculated from the previous generation (or the initial state assumed to be in HWE) in the Hardy-Weinberg equations.
Using p = 0.553 and q = 0.447 (from Generation 1):
- Expected frequency of AA (p^2): (0.553)^2 ext{ (approximately } 0.305)
- Expected frequency of Aa (2pq): 2 imes 0.553 imes 0.447 ext{ (approximately } 0.494)
- Expected frequency of aa (q^2): (0.447)^2 ext{ (approximately } 0.199)

Compare the observed genotype frequencies from Step 2 with the expected genotype frequencies from Step 3.
Question: Are the expected genotype frequencies significantly different from our observed genotype frequencies? (In this example, the observed were 0.30, 0.50, 0.20 and expected were approximately 0.305, 0.494, 0.199).
Conclusion for Population 1: No, the observed frequencies are not significantly different from the expected frequencies.
- Therefore, we can conclude that this population is not evolving at this locus.
- This means the allele frequencies are stable and not changing from one generation to the next.

Scenario: Given new numbers for the number of individuals at each genotype for Population 2, Generation 2.
Task: Determine if Population 2, Generation 2, is in Hardy-Weinberg Equilibrium.
Steps to follow:
1. Calculate allele frequencies (p and q) from Population 2, Generation 1 (if provided, or assume current observed values if comparing against themselves).
2. Calculate observed genotype frequencies for Population 2, Generation 2.
3. Calculate expected genotype frequencies for Population 2, Generation 2 (using p and q from step 1).
4. Compare observed and expected frequencies to draw a conclusion about evolution.
Reminder: When calculating allele frequencies from individual counts, remember that the total number of alleles is 2 imes ext{number of individuals}.
- If there are 200 individuals, there are 400 alleles.