If a woman has FEVR and is affected by the disease, what are the chances of her passing the disease to future generations. I read somewhere from a credible source like NIH(I couldn't find the link again, I am attaching another https://rarediseases.info.nih.gov/diseases/1613/familial-exudative-vitreoretinopathy#:~:text=Familial%20exudative%20vitreoretinopathy%20(FEVR)%20is,the%20blood%20supply%20to%20retina.), that the babies that the woman conceives could be affected by or carrying the FEVR causing genes, or may neither be carrying nor affected by them. Depending on the type of inheritance, which are autosomal dominant, autosomal recessive, or X-linked recessive, the chances of having a normal baby vary. The chances of having a healthy kid for autosomal dominant and autosomal recessive are 50% and 25%, respectively ( I am intentionally excluding the possibility of X-linked recession).

Therefore, if the inheritance is autosomal dominant, then, there is a 50% chance that the baby is healthy and not carrying the genes. And the successive generations created by the baby will not have the gene causing FEVR. Is my understanding correct?

  • 1
    $\begingroup$ Are you asking about the mendelian inheritance calculations? $\endgroup$
    – Roger V.
    Commented Dec 24, 2020 at 6:35

1 Answer 1


The answer depends on (at least):

  • whether the woman is affected by the condition
  • whether the allele is autosomal dominant or autosomal recessive (excluding X-linked recession)
  • the frequency of the diseased alleles in the population
  • the presence or not of inbreeding

Below I assume no inbreeding.

  1. The woman is affected and the inheritance is autosomal recessive. This means that the woman carries two disieased alleles and with certainty passes one of them to her children (i.e., with probability $1$). Whether the children will be affected by the condition depends on whether they get a second diseased allele from the other parent. Assuming that the frequency of the diseased allele in the population is $p$, this means that the probability of a child inheriting the disease is $p$.
  2. The woman is not affected and the inheritance is autosomal recessive. The woman carries only one diseased allele, and passes it with probability $1/2$ to her child. The probability that a child is affected, i.e. that s/he has two diseased alleles is $p/2$.
  3. The woman is not affected and the inheritance is autosomal dominant. In this case the woman carries no diseased allele, so the children are not affected.
  4. The woman is affected and the inheritance is autosomal dominant. The woman carries at least one diseased allele, and could be carrying the second with probability $p$. In the former case she passes it to her child with probability $1/2$, in the latter with probability $1$, so the probability that the child inherits the diseased allele is $1/2 + p$. This is also the probability that the child is affected by the condition, since they will be affected regardless of the allele passed by the other parent.

Assuming very small probability $p$, case 4 gives us the probability $50%$, whereas the probabily in cases 1 and 2 is much lower. The numbers given in the question are calculated assuming Mendelian experiments, which are a heavy case of inbreeding (and even then not quite correct).

John H.Gellespie, Population genetics
Matthew Hamilton, Population genetics

  • 1
    $\begingroup$ The above assumes complete penetrance; many disease show only partial penetrance, where some individuals who carry the disease allele(s) show no sign of disease. That greatly complicates calculations, significantly changing cases 2 and 3 above. $\endgroup$
    – Armand
    Commented Sep 21, 2021 at 0:37

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