MP Board · Class 12 · Biology · EvolutionIn a stable population in Hardy-Weinberg equilibrium, 16% of the individuals display a recessive trait ($aa$). Calculate: The allele frequencies of the recessive allele ($a$) and the dominant allele ($A$). The percentage of the population expected to be heterozygous ($Aa$). The percentage of the population expected to be homozygous dominant ($AA$).

Given Data:

• Percentage of homozygous recessive individuals ($aa$) = 16% = 0.16

Hardy-Weinberg Principle Equations:

1. $p + q = 1$
2. $p^2 + 2pq + q^2 = 1$ \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 ($aa$)

Step-by-Step Solution:

1. Frequency of alleles ($a$ and $A$):

• Given, $q^2 = 0.16$

• $q = \sqrt{0.16} = 0.4$

• Therefore, the frequency of recessive allele ($a$) is 0.4.

• Using $p + q = 1$:

• $p = 1 - q = 1 - 0.4 = 0.6$

• Therefore, the frequency of dominant allele ($A$) is 0.6.

2. Percentage of heterozygous individuals ($Aa$):

• Frequency of $Aa = 2pq$
• $2pq = 2 \times 0.6 \times 0.4 = 0.48$
• Percentage = $0.48 \times 100 =$ 48%

3. Percentage of homozygous dominant individuals ($AA$):

• Frequency of $AA = p^2$
• $p^2 = (0.6)^2 = 0.36$
• Percentage = $0.36 \times 100 =$ 36%