A genetic locus can have two alleles, A and B. Since each person has two copies (one from each chromosome), there are three genotypes: homozy- gous AA, BB, and heterozygous AB. Note that since we usually do not know the descent of each allele (father or mother), AB and BA are the same genotype. Suppose the frequency of allele A in a population is 90%. Suppose each individual has two independent alleles at this locus. What is the probability that a person’s phenotype is AA, if we know that he has at least one copy of the A allele?
My calculation is as follows:
$$\frac{P(AA)}{P(AA)+P(AB)} = \frac{0.9 \cdot 0.9}{0.9 \cdot 0.9 + 0.9 \cdot 0.1} = 0.9$$
Is this correct? Thanks for help!
I see one answer says that "AB" stands for "AB" and "BA", but "AB" and "BA" are the same genotype, why should we consider "AB" in two different ways and double the probability?
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$\begingroup$Not quite – you didn't account for the fact that AB occurs in two different ways – "AB" stands for AB and BA combined, and the probability for this is $0.9\cdot0.1+0.1\cdot0.9=2\cdot0.9\cdot0.1$.
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