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Hereditary characteristics are determined by pairs of genes. A gene pair for a particular characteristic is transmitted from parents to offspring by choosing randomly one gene from the mother’s pair and one genera from the father’s pair. Each gene may have several types or alleles. For example human beings have one allele denoted by B for brown eyes and one allele denoted by b for blue eyes. A person with the allele pair BB has brown eyes, with bb has blue eyes. A person with the pair Bb ( or bB which is the same thing) will also have brown eyes and the allele B is said to be dominant.

A brown eyed woman has brown eyed parents, both known to be Bb. She and a blue-eyed man have a child. Given the child is brown eyed what is the chance the woman carries the allele b?

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If woman's parents are $Bb$, then she may be $BB$, $Bb$, $bB$ or $bb$ with $25\%$ chance for each genotype. As we know that her phenotype is "brown eyes", then the last one is not possible and we have equal chances of $BB$, $bB$ and $Bb$ for this woman. As $bB$ and $Bb$ are the same, we can say that she's $1/3$ homozygous and $2/3$ heterozygous. Blue-eyed man is definitely $bb$, while brown-eyed child can be $Bb$ or $BB$. As the child can't be $BB$ with $bb$ father, that child is definitely $Bb$ (or $bB$, let's consider them as the same).
So we have the following:
- mother is $1/3$ chance $BB$ and $2/3$ chance $Bb$
- father is $100\%$ $bb$
- child is $100\%$ $Bb$

$BB$ mother with $bb$ father will always have $Bb$ child, while $Bb$ mother with $bb$ father only in half of cases. So, $1/3$ for $BB$ and $2/3 * 1/2 = 1/3$ for $Bb$. Normalizing the weights of the events to make the sum $= 1$, we have that chance the woman is $Bb$ is $1/2$.

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  • $\begingroup$ Very nicely done. Thanks :D $\endgroup$ – Namch96 Feb 4 '15 at 15:46

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