Answer: 0.18
Explanation:
For the alleles, the percentage distribution of each is 'A' (90% = 0.9)
While 'a' (10% = 0.1)
Hence, 0.9 and 0.1 are the respective frequencies of each allele
Now, apply Hardy-Weinberg Equilibrium equation, where heterozygotes are represented by the 2pq term.
Therefore, the number of heterozygous individuals (Aa) is equal to 2pq which equals
2 × 0.9 × 0.1 = 0.18
Thus, the frequency of heterozygote is 0.18, while the percentage distribution in the population is 18%
1st image I think would be the answer
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