• Hardy-Weinberg Equilibrium Genotype Frequencies:
• For a diallelic gene with allele frequencies $p$ (allele $A$) and $q$ (allele $a$) such that $p + q = 1$:
• Heterozygote frequency: $f(Aa) = 2pq$
• Recessive homozygote frequency: $f(aa) = q^2$
• Given condition: The heterozygous genotype is twice as frequent as the homozygous recessive genotype:

• Dividing both sides by $2q$ (for $q \neq 0$):

• Since $p + q = 1$ and $p = q$:

• Conclusion: The required allelic frequencies are p = 0.5; q = 0.5 (Option 4).

Correct option: (4) p=05;q=05