One of my friend was saying that her aunt (father's sibling sister) was having polydactyly but none of her cousins (her aunt has two daughters) is having it.

What is the chance that my friend's children inherit polydactyly?

(considering that her partner nor anyone in his generation exhibit polydactyly)

  • 1
    $\begingroup$ if polydactyly is autosomal dominant (Wikipedia), then effectively zero. $\endgroup$
    – Ben Bolker
    Dec 29, 2018 at 21:53

1 Answer 1


Polydactyly can be dominant or recessive. that is

  1. A single copy of the mutation/s involved result in the phenotype vs
  2. Both copies of the gene must be mutated to result in the phenotype

When two random people who don't have polydactyly nor a family history of polydactyly have a child, there is around 1/548 (0.18%) chance that child will be polydactyly because the parents were carriers without knowing it, and a 1/11.7 (8.54%) chance the child is a carrier of a single copy

If the aunt's polydactyly is Dominant, then your friend's father hasn't inherited it (he isnt polydactyly) and just has the normal population chance of being a carrier.

In the case of Dominant polydactyly, assuming your friend has a partner with no history of polydactyly, your friend and her partner have the same chance as average of a polydactyly child (1/548)

If the aunt's polydactyly is recessive, meaning she has two copies one from each parent (you friend's grandparents), there is a 1/130 (0.76%)chance your friend's child will have polydactyly (4.2 times more likely), and a 1/4.5 (21.4%) chance of being a carrier (2.6 times more likely)

I'll quickly go through the working/ if you're interested:

Given that the grandparents are both carriers (needed for the aunt to have two copies), then your friend's father has a 2/3 chance of being a carrier of a single copy (1/3 chance no copies, 0/3 chance having two copies because he isn't polydactyly).

Your friend's mother has the usual 1/11.7 chance of being a carrier.

The chance of your friend inheriting a single copy is (chance passes from father) + (chance passes from mother) - (chance of getting two copies)

=0.6600/2 + 0.0854/2 - (0.33*0.0427)

=0.3727 - 0.0140

= 0.3586

35.9% chance your friend carries one copy

Your friend has 1/2.8 chance of carrying and 1/5.6 of passing it on

Friend's partner has the 1/11.7 chance of being a carrier and the 1/23.4 chance of passing it on

Therefore their child has

=0.1793*0.0427 (chance friend passes on * chance partner passes on)


=1/130 chance of having two copies, having polydactyly

=0.1793 + 0.0427 -(0.1793*0.0427) (chance any copies - chance two copies)


=1/4.7 chance of one copy a 2.5 times increase risk

  • 1
    $\begingroup$ That's a really detailed answer. Good one and thanks for simplifying the confusing parts. But can you add any reference? It will help me accept your answer. $\endgroup$
    – user 33690
    Dec 31, 2018 at 18:16

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