I was doing some pedigree work and couldn't figure out why my answer is wrong. This chart is supposed to be an autosomal Dominant method of inheritance, but I am finding an autosomal recessive pattern that fits? If anyone can point out my mistake, that would be appreciated. I have labelled my attempt above each member, where aa = disease phenotype and Aa or AA is a normal phenotype. The disease pattern describes Marfan syndrome (autosomal dominant). Any help or pointers would be appreciated.

I decided not to label the pt who has a *'s family, because there are several possibilities that don't affect the answer.

Edit: I also found an autosomal dominant pattern that fits, but the issue is with the autosomal recessive pattern, so I am only posting that one.

My attempt

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    $\begingroup$ Is it reasonable to assume that everybody that enters/"marries into" this family tree happens to be a carrier for the same recessive trait? $\endgroup$
    – Bryan Krause
    Commented Feb 26 at 0:57

1 Answer 1


I do not see any "error" in the logic of your pedigree.

However, pedigree analysis is not a logic puzzle; it's a practical tool. There are additional logical solutions you have not come up with. For example, every affected individual in this tree could result from a de novo somatic mutation with no inheritance at all. Or the phenotype could be not genetic at all: the observed pedigree pattern might well apply to "people who have been in a serious car accident". Or, it could be a polygenic trait that is not easily tracked in pedigrees even if there is some underlying inheritance.

You're expected to have some additional assumptions that are met to even bother undertaking a pedigree analysis, and one of those likely assumptions is that the disease is rare. That is, it's rare enough to rule out more ordinary polygenic traits and make you think there's a single gene involved. In that case, it is not reasonable to assume that everyone entering the pedigree from outside (e.g. marrying in/breeding in) happens to be a carrier of the disease.

It looks like that's what was necessary to make this tree work: besides the initial aa/Aa pairing at the top, there are 4 other individuals that come from outside and pair with someone who is a descendent of that pairing (there are actually 5 in a sense, if you include both of *'s parents, but we can reduce it to just * and count them as #4). You've marked all 4 of these as heterozygous carriers for this pedigree to work. It would be extremely unlikely for this to happen with a rare disease.

See also the concept of Occam's razor.

  • $\begingroup$ Bryan already answered in a very exhaustive and elegant way. The fact is that most textbooks and teachers, when dealing with recessive traits, assume they are severe diseases, for which it is reasonable to expect that they are rare in the population. This will also make the recessive allele rare in the population, and the likelihood that 4 married-in all carry the disease allele is so low to be neglectable. $\endgroup$ Commented Mar 2 at 14:11

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