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Let p = the frequency of the dominant allele and q = the frequency of the recessive allele. The following statement should therefore be true:
p + q = 1
In other words, the fraction of dominant alleles plus the fraction of recessive alleles should account for ALL of the alleles (1 = 100%).
Now if we assume the population is in genetic equilibrium, then the fraction of individuals who have the BB genotype should be p², and the fraction of individuals who have the bb genotype is q². The fraction of individuals who are heterozygous (Bb) is given by the expression 2pq. So:
p² + 2pq + q² = 1
In other words, the homozygous dominant individuals plus the heterozygous dominant individuals plus the homozygous recessive individuals should account for ALL invidivuals.
Now we know that 25% of the population is bb. The other 75% is BB + Bb. So:
p² + 2pq = 0.75
q² = 0.25
We can't solve the first equation yet, but we can solve for q:
q = square root of 0.25 = 0.5
So one-half of the population's alleles is recessive.
The other half, then, must be dominant. p = 0.5 as well.
Now that we know the values of p and q, we can solve for p² (the frequency of homo. dominant individuals) and 2pq (the hetero. dominant individuals)
p² = (0.5)² = 0.25
2pq = 2(0.5)(0.5) = 0.5
There you have it: one quarter of the population is BB, one half is Bb, and one quarter is bb. A perfect 1:2:1 distribution...
I hope that helps. Good luck!
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