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.



(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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