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Archiver > GENEALOGY-DNA > 2002-01 > 1011047121


From:
Subject: Re: [DNA] Atlantic Modal Haplotype
Date: Mon, 14 Jan 2002 20:25:21 -0200


John, Obviously the 5 AMH markers are influenced by some important
OTHER factors in order to conclude that they "cluster". As you show,
that
makes it about 6 times harder (4% to 25% for a 4 out of 5 match)to claim
that
whatever differences are seen, in each Markers comparison to the AMH,
exist INDEPENDENTLY of each other. Could the notion that each marker
is "equal" to any other be an erroneous postulate in attempts to
generalize
to either the Group or to the Individual? If erroneous, then how to
correct?
Dick Matteson

On Mon, 14 Jan 2002 16:04 EST "John F. Chandler"
<> writes:
> Ann wrote:
> > I'm not as disturbed by the numbers as you are, I guess. The AMH
> is based on
> > only 6 markers, and every marker you add reduces the frequency of
> exact
> > matches. The Y-STR database at www.ystr.org/europe doesn't include
> DYS388,
> > but if you plug in the values for the other markers, the AMH
> frequency in the
> > European database is about 10%.
-----------------------------------------------------
> True, but if you add up the totals for not only the AMH itself, but
> also
> all the off-by-one variations, it comes to about 25% in all (2178
> out of
> 8592). Obviously, things are (or at least should be) a lot better
> with
> 12 markers instead of 5, but even then the statistics won't be
> ideal
> because of the strong clustering on these 5. For comparison,
> suppose
> each of the five loci had an independent random value chosen within
> a
> range of three adjacent values (i.e., the median value plus or
> minus
> one). The probability of getting an exact match to the median for
> all
> five would be (1/3)^5 = 0.4%, and the probability of getting a 4/5
> match
> with a one-step offset on the fifth would be (1/3)^4*(2/3)*5 = 4%.
> If
> you spread each locus out over 5 adjacent values, then the
> probabilities
> drop to 0.03% and 0.3%, respectively.
>
> John Chandler
>
>
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