Hardy-Weinberg equilibrium for X-linked recessive genes:

• Males are hemizygous ($X^a Y$); hence the incidence of an X-linked recessive disorder in males directly equals the allele frequency $q$: $q = 1/100 = 0.01$.
• Females have two X chromosomes ($X^a X^a$); hence the expected frequency of affected homozygous females is $q^2$.
• Calculating $q^2$: $q^2 = (0.01)^2 = 0.0001 = 1 / 10000$.
• Thus, the expected incidence in females is $1\text{ in } 10,000$.

Genotype frequency calculation in a population:

• Carrying two specific distinct alleles ($A_1$ and $A_2$) corresponds to being heterozygous for these two alleles ($A_1 A_2$).
• Under Hardy-Weinberg equilibrium, the expected frequency of a heterozygote with allele frequencies $p$ and $q$ is $2pq$.
• Given $p = 0.2$ and $q = 0.2$: $\text{Frequency} = 2pq = 2 \times 0.2 \times 0.2 = 0.08$.
• Therefore, the probability is $0.08$.