Answer:
980,100
Explanation:
let the recessive condition (aa) be represented as q² which is the genotypic frequency = 100/1,000,000 = 0.0001
allelic frequency of a = q = √q² = √0.0001 = 0.01. thus q = 0.01
From the formula p + q = 1 where p is the allelic frequency of A. 
since q = 0.01 
p = 1 - 0.01 = 0.99. The allelic frequency of A (p) = 0.99
p² = 0.99² = 0.9801. Genotypic frequency of AA= 0.9801 
= 0.9801 x 1,000,000 = 980,100 individuals with AA (homozygous normal)
For 2pq genotypic frequency for hetrozygous (Aa). 
Using the formula p² + 2pq + q² = 1.  Since p² = 0.9801 and q² = 0.0001
    2pq = 1 - (p² + q²)
 = 1 - (0.9801 + 0.0001)
 = 1 - (0.9802)
 = 0.0198 = 0.0198 x 1,000,000 = 19,800 individuals with Aa 
 
        
             
        
        
        
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