MP Board · Class 12 · Biology · EvolutionIn a random mating population in Hardy-Weinberg equilibrium, the frequency of a recessive autosomal genetic disease is 1 in 10,000 ($0.0001$). Calculate: The frequency of the recessive allele ($q$) The frequency of the dominant allele ($p$) The frequency of carriers (heterozygous individuals) in the population.\nShow all steps clearly.

Step-by-Step Solution:

Given Data:

• Frequency of the recessive disease (homozygous recessive individuals, $aa$) = $q^2 = \frac{1}{10000} = 0.0001$

1. Calculation of the frequency of the recessive allele ($q$): $$q^2 = 0.0001$$\nTaking the square root on both sides: $$q = \sqrt{0.0001}$$ $$q = 0.01$$ Answer 1: The frequency of the recessive allele ($q$) is $0.01$ (or $1%$).

2. Calculation of the frequency of the dominant allele ($p$):\nAccording to the Hardy-Weinberg principle: $$p + q = 1$$\nSubstitute the value of $q$: $$p + 0.01 = 1$$ $$p = 1 - 0.01$$ $$p = 0.99$$ Answer 2: The frequency of the dominant allele ($p$) is $0.99$ (or $99%$).

3. Calculation of the frequency of carriers (heterozygous individuals):\nIn Hardy-Weinberg equilibrium, heterozygous carriers are represented by the term $2pq$. $$2pq = 2 \times p \times q$$\nSubstitute the values of $p$ and $q$: $$2pq = 2 \times 0.99 \times 0.01$$ $$2pq = 2 \times 0.0099$$ $$2pq = 0.0198$$ Answer 3: The frequency of carriers (heterozygous individuals) in the population is $0.0198$ (or approximately $1.98%$).