CBSE · Class 12 · Biology · EvolutionState and explain the Hardy-Weinberg Principle. List the five factors that affect the Hardy-Weinberg equilibrium. Numerical Problem:\nIn a population of 1000 individuals, 360 belong to genotype AA, 480 to Aa, and 160 to aa. Calculate the allele frequencies of allele 'A' and allele 'a'. Verify whether this population is in Hardy-Weinberg equilibrium.

Hardy-Weinberg Principle\nThe Hardy-Weinberg Principle states that allele frequencies in a population are stable and remain constant from generation to generation in the absence of evolutionary influences. The total gene pool (total genes and their alleles in a population) remains constant. This is known as genetic equilibrium or Hardy-Weinberg equilibrium.

\nThe sum total of all the allelic frequencies in a population is equal to 1: $$p^2 + 2pq + q^2 = 1$$\nWhere:

• $p$ = frequency of dominant allele 'A'
• $q$ = frequency of 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)

Factors Affecting Hardy-Weinberg Equilibrium\nFive main factors are known to affect Hardy-Weinberg equilibrium and cause deviation from evolutionary stability:

1. Gene Migration or Gene Flow: When individuals migrate to another population or new individuals enter a population, gene frequencies change in both the original and the new population.
2. Genetic Drift: Random changes in allele frequencies occurring purely by chance, especially in small populations.
3. Mutation: Random, heritable changes in DNA sequences that introduce new alleles into a gene pool.
4. Genetic Recombination: Recombination of genes during meiosis and crossing over leads to new combinations of alleles in offspring.
5. Natural Selection: A process in which heritable traits that increase survival and reproduction become more common in successive generations.

Numerical Solution:

Given Data:

• Total population size ($N$) = $1000$
• Number of $AA$ individuals = $360$
• Number of $Aa$ individuals = $480$
• Number of $aa$ individuals = $160$

Step 1: Calculate Genotype Frequencies

• Frequency of genotype $AA$ ($p^2$) = $\frac{360}{1000} = 0.36$
• Frequency of genotype $Aa$ ($2pq$) = $\frac{480}{1000} = 0.48$
• Frequency of genotype $aa$ ($q^2$) = $\frac{160}{1000} = 0.16$

Step 2: Calculate Allele Frequencies ($p$ and $q$)

• Allele frequency of 'A' ($p$): $$p = \sqrt{p^2} = \sqrt{0.36} = 0.6$$ (Alternatively: $p = \text{Frequency of } AA + \frac{1}{2}(\text{Frequency of } Aa) = 0.36 + 0.24 = 0.60$)

• Allele frequency of 'a' ($q$): $$q = \sqrt{q^2} = \sqrt{0.16} = 0.4$$ (Alternatively: $q = 1 - p = 1 - 0.6 = 0.4$)

Step 3: Verification of Equilibrium

• Check if $p + q = 1$: $$0.6 + 0.4 = 1.0$$
• Check if $p^2 + 2pq + q^2 = 1$: $$(0.6)^2 + 2(0.6)(0.4) + (0.4)^2 = 0.36 + 0.48 + 0.16 = 1.0$$

Conclusion:\nThe gene frequency of allele 'A' is 0.6 and allele 'a' is 0.4. Since the observed genotype frequencies strictly follow the equation $p^2 + 2pq + q^2 = 1$, the population is in Hardy-Weinberg equilibrium.