If a woman has a single male son affected by an X-linked recessive condition, then the probability that she is a carrier can be calculated if she has normal brothers and other sons who are all normal.
Assumptions:
(1) mutation rates for males and females are equal
(2) reproductive fitness for affected males is zero (f = 0)
(3) coefficient of µ for the female carrier = 4
(4) the woman only has 1 affected son
posterior probability that the woman is a carrier for an X-linked recessive condition =
= ((2 ^ (number of normal brothers)) + 1) / ((2 ^ ((number of normal sons) + (number of normal brothers))) + (2 ^ (number of normal sons)) + 1)
The derivation is given in Table A.5 in Bridge (1997).
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