MP Board · Class 12 · Biology · EvolutionState the Hardy-Weinberg Principle. In a stable, randomly-mating population of 1000 individuals, 360 individuals display the recessive phenotype caused by a homozygous recessive genotype ($aa$). Assuming the population is in Hardy-Weinberg equilibrium, calculate: The frequency of the recessive allele ($a$) and the dominant allele ($A$). The expected number of homozygous dominant ($AA$) and heterozygous ($Aa$) individuals in the population. State three key factors that can disturb Hardy-Weinberg equilibrium in a natural population.

1. Hardy-Weinberg Principle\nThe Hardy-Weinberg principle states that allele frequencies in a population remain stable and constant from generation to generation, provided there are no evolutionary influences such as mutation, gene flow, genetic drift, natural selection, or non-random mating. This equilibrium is mathematically represented as:

$$p^2 + 2pq + q^2 = 1$$\nand $$p + q = 1$$\nWhere:

• $p$ = frequency of the dominant allele ($A$)
• $q$ = frequency of the recessive allele ($a$)
• $p^2$ = frequency of homozygous dominant individuals ($AA$)
• $2pq$ = frequency of heterozygous individuals ($Aa$)
• $q^2$ = frequency of homozygous recessive individuals ($aa$)

2. Step-by-Step Calculation

Given:

• Total population size ($N$) = $1000$
• Number of homozygous recessive individuals ($aa$) = $360$

Step A: Calculate allele frequencies ($q$ and $p$)

1. Frequency of homozygous recessive genotype ($q^2$): $$q^2 = \frac{\text{Number of } aa \text{ individuals}}{\text{Total population}} = \frac{360}{1000} = 0.36$$

2. Frequency of recessive allele ($a$ or $q$): $$q = \sqrt{0.36} = 0.6$$

3. Frequency of dominant allele ($A$ or $p$): $$p = 1 - q = 1 - 0.6 = 0.4$$

Step B: Calculate expected genotype frequencies and individual numbers

1. Homozygous dominant individuals ($AA$):

   • Frequency ($p^2$) = $(0.4)^2 = 0.16$
   • Expected number = $p^2 \times N = 0.16 \times 1000 = 160 \text{ individuals}$

2. Heterozygous individuals ($Aa$):

   • Frequency ($2pq$) = $2 \times 0.4 \times 0.6 = 0.48$
   • Expected number = $2pq \times N = 0.48 \times 1000 = 480 \text{ individuals}$

Verification:

• Total individuals = $160 (AA) + 480 (Aa) + 360 (aa) = 1000$

3. Factors Disturbing Hardy-Weinberg Equilibrium

1. Gene Migration / Gene Flow: Movement of individuals into (immigration) or out of (emigration) a population alters allele frequencies.
2. Genetic Drift: Random changes in allele frequency occurring purely by chance, especially in small populations.
3. Natural Selection: Differential survival and reproduction of genotypes favoring advantageous alleles.