Archiver > GENEALOGY-DNA > 2005-06 > 1120062225

From: "William Hurst" <>
Subject: Re: [DNA] R1b match in Fortineux Project?
Date: Wed, 29 Jun 2005 12:23:45 -0400
References: <006801c57c46$b71e7a30$7302a8c0@YOURF8387228BF> <002701c57c96$f25e1d80$7302a8c0@YOURF8387228BF>

Phil wrote:

> Anybody out there with R1b experience?

Yep, I've apparently been one all my life.

> ....My two participants match on 22
> markers. The remaining three markers are one-step mismatches:
> Kit 33862 Kit 34434
> DYS390 23 24
> DYS392 13 14
> DYS464d 17 18
> While we can't deduce the ancestral values for these three markers without
> a
> third participant, it seems that any combination would mean that one
> participant has a genetic distance of one and the other participant has a
> genetic distance of two, which would seem reasonable for two men with a
> common ancestor born in about 1650. For example, if the Jonas Fortineux
> ancestral haplotype was 24.13.18 for these three markers, kit #33862 has a
> genetic distance of two and kit #34434 has a genetic distance of one.
> Is it reasonable to conclude that these two men, with variants of the same
> surname and a paper trail to Jonas Fortineux (b ca1650) are in fact
> genetically descended from Jonas Fortineux? Or is it a stretch to
> "triangulate" when one is missing the third corner of a triangle. In this
> case, might that third corner be the similar surname and the paper trail.
> I
> would be grateful for anyone with some knowledge of R1b to offer an
> opinion.
> Thanks,

I don't think having three participants somehow magically allows you to
determine an ancestral haplotype; it just provides a better estimate. In the
original concept of triangulation, the three corners are the ancestor and
the descendants of two different sons. If the two descendants have an exact
match, then their haplotype was the haplotype of the ancestor - barring very
rare parallel mutations. After the concept was introduced here by me, it was
extended to estimating ancestral haplotypes when there were no exact
matches. Note that you really need three haplotypes from descendants of
three different sons of the ancestor. Then it's a matter of taking the best
two-out-of-three for each mutation. Lots of luck if each of the three has a
different value on a particular marker. If you have this situation, or yours
where there are only two haplotypes, you might have to assume that the modal
marker value - from, say, the WAMF - is your ancestral marker value.

Having the same surname and a paper trail is very important. I wouldn't
worry about a genetic distance of two based on an ancestor from 1650. In our
group, we have only one man of 15 with a distance of two from the ancestor,
but he has a genetic distance of three from his closest cousin. We are lucky
in that we have perfect matches between between several descendants of
several men all leading back to a Hurst born about 1675. So far.

I submit that the first man with a haplotype determined by triangulation was
Absalom Hurst (1750-1830). A 4th cousin, twice removed, and I have excellent
paper trails back to two of his sons. Since we have a 37/37 match, it
occurred to me one day that Absalom had that same haplotype and that
"triangulation" would be a good word to describe how I could prove this.
(Note that list member cousin Ernie Hurst is probably descended from a third
son of Absalom and shares the ancestral haplotype.) From tests by
descendants as far apart as 7th cousin, we now know that Absalom's
grandfather had the same haplotype. We are still working on the proof that
he was the John Hurst who died in Stafford County, VA, in 1747.

Bill Hurst

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