Answer:
A) 198
Explanation:
Solution:
frequency of aa = q2 = 1/10,000 = 0.0001, q = 0.01
number of aa = 0.0001 x 10,000 = 1 individual
p + q = 1, p = 0.99
frequency of AA = p2 = 0.9801
number of AA = 0.9801 x 10,000 = 9,801 individuals
Aa = 2pq = 2 x 0.9801 x 0.01 = 0.0198 or 198 individuals
A population of 10,000 mussels, how many individuals would be heterozygous and contain the recessive gene that might potentially help rid the Great Lakes of this invasive species is 198.
frequency of aa = q2 = 1/10,000 = 0.0001, q = 0.01
number of aa = 0.0001 x 10,000 = 1 individual
p + q = 1, p = 0.99
frequency of AA = p2 = 0.9801
number of AA = 0.9801 x 10,000 = 9,801 individuals
Aa = 2pq = 2 x 0.9801 x 0.01 = 0.0198 or 198 individuals
Thus, Aa = 2pq = 2 x 0.9801 x 0.01 = 0.0198 or 198 individuals.
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