CBSE · Class 12 · Biology · EvolutionQuestion 1 (a) State the Hardy-Weinberg Principle and write its mathematical equation. (b) Solve the following numerical problem:\nIn a random mating population of $1000$ individuals, $360$ belong to genotype $AA$, $480$ to $Aa$, and $160$ to $aa$. Calculate the frequency of allele $A$ and allele $a$. Verify if the population is in Hardy-Weinberg equilibrium.

Part (a): Hardy-Weinberg Principle

• Definition: The 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 (such as mutation, gene flow, genetic drift, natural selection, and non-random mating). The gene pool (total genes and their alleles in a population) remains constant. This is known as genetic equilibrium.
• Mathematical Equation: If $p$ represents the frequency of dominant allele ($A$) and $q$ represents the frequency of recessive allele ($a$): $$p + q = 1$$ The binomial expansion for genotypic frequencies is: $$(p + q)^2 = p^2 + 2pq + q^2 = 1$$ Where:
  • $p^2$ = frequency of homozygous dominant genotype ($AA$)
  • $2pq$ = frequency of heterozygous genotype ($Aa$)
  • $q^2$ = frequency of homozygous recessive genotype ($aa$)

Part (b): Step-by-Step 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 Observed Genotypic Frequencies

• Observed frequency of $AA$ ($p^2_{obs}$) = $\frac{360}{1000} = 0.36$
• Observed frequency of $Aa$ ($2pq_{obs}$) = $\frac{480}{1000} = 0.48$
• Observed frequency of $aa$ ($q^2_{obs}$) = $\frac{160}{1000} = 0.16$

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

• Frequency of allele $A$ ($p$): $$p = \text{Freq}(AA) + \frac{1}{2} \times \text{Freq}(Aa)$$ $$p = 0.36 + \frac{1}{2}(0.48) = 0.36 + 0.24 = 0.60$$

• Frequency of allele $a$ ($q$): $$q = \text{Freq}(aa) + \frac{1}{2} \times \text{Freq}(Aa)$$ $$q = 0.16 + \frac{1}{2}(0.48) = 0.16 + 0.24 = 0.40$$

(Check: $p + q = 0.60 + 0.40 = 1.00$)

Step 3: Calculate Expected Genotypic Frequencies\nUsing Hardy-Weinberg equation $(p+q)^2 = p^2 + 2pq + q^2 = 1$:

• Expected $AA$ frequency ($p^2$) = $(0.60)^2 = 0.36$
• Expected $Aa$ frequency ($2pq$) = $2 \times 0.60 \times 0.40 = 0.48$
• Expected $aa$ frequency ($q^2$) = $(0.40)^2 = 0.16$

Conclusion:\nSince the observed genotypic frequencies ($AA = 0.36$, $Aa = 0.48$, $aa = 0.16$) are exactly equal to the expected genotypic frequencies calculated using Hardy-Weinberg equilibrium equation, the population is in Hardy-Weinberg equilibrium.