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From: "Alister John Marsh" <>
Subject: Re: [DNA] Variance Assessment of R:U106 DYS425Null Cluster
Date: Tue, 9 Feb 2010 19:24:18 +1300
References: <mailman.3478.1265650803.2099.genealogy-dna@rootsweb.com><E5ABF747646147CF8780B5FD8508EA59@anatoldesktop>
In-Reply-To: <E5ABF747646147CF8780B5FD8508EA59@anatoldesktop>


Anatole,



You gave this example below in your recent posting (Have added letters A to
H to identify individual haplotypes)...

>>>>>>>

Now, let ME give you a couple of examples, out of many. A reader sent me a
25-marker haplotype set of eight relatives in Britain. He asked me to
determine when a common ancestor lived, however, did not disclose the actual
date. The list was as follows:





A 13 25 14 10 11 14 12 12 10 13 11 30 15 9 10 11 11 23 14 20 35 15 15 15 16

B 13 25 15 10 11 14 12 12 10 13 11 29 15 9 10 11 11 23 14 20 35 15 15 15 16

C 13 25 15 10 11 14 12 12 10 13 11 30 15 9 9 11 11 23 14 20 35 15 15 15 15

D 13 25 15 10 11 14 12 12 10 13 11 30 15 9 10 11 11 23 14 20 35 15 15 15 16

E 13 25 15 10 11 14 12 12 10 13 11 30 15 9 10 11 11 23 14 20 35 15 15 15 16

F 13 25 15 10 11 14 12 12 10 13 11 30 15 9 10 11 11 23 14 20 35 15 15 15 16

G 13 25 15 10 11 14 12 12 10 13 11 30 15 9 10 11 11 23 14 20 35 15 15 15 16

H 13 25 15 10 11 14 12 12 10 13 11 30 15 9 10 11 11 23 14 20 35 15 15 15 16

Clearly, the base, ancestral haplotype is as follows:



13 25 15 10 11 14 12 12 10 13 11 30 15 9 10 11 11 23 14 20 35 15 15 15 16



All eight haplotypes have three mutations per 200 alleles. It gives

3/8/0.046 = 8 generation from a common ancestor. At the same time the series
contains five base haplotypes, which gives ln(8/5)/0.046 = 10 generations.

It gives the average value of 9 generation, plus-minus some margin of error.


However, a formula (given in my publication I have referred to) shows that
for a 200-marker series and three mutations in it, for a fully asymmetrical
mutations (which the series shows) a standard deviation theoretically equals
to 57.7%. At the 68% confidence level ("one sigma") we obtain that there
would be 9±3 generations, and at the 95% confidence level ("two sigma")
there would be 9±5 generations to the common ancestor. Therefore, the common
ancestor lived in 1784±75 (68% confidence), or in 1784±125 (95% confidence)
years ago, and, since he was born some 25 years earlier, it gives his birth
year around 1759, give or take a century. In fact, as I was later informed,
Robert, the common ancestor of all the eight individuals, was born in 1767.

<<<<<<<



I am not a mathematician, so you may be able to correct me, but would not
the history of this group have a bearing on the expected number of
mutations? What I mean is, suppose B, D, E, F, G, H were brothers (who have
identical haplotypes), and the other two A and C of unknown relationship to
the 5 brothers? Would that in effect mean that comparing haplotypes A, B,
C, would be a more meaningful way to assess age since common ancestor using
your formula for this group?



Alternatively, suppose that in reality, C was exactly the same as the
ancestral haplotype, it would be less easy to estimate how many mutations
had taken place in the group, as some lines may have had parallel mutations.



Vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv



I have an example for you to consider. Can you make anything of the 11
haplotypes of 37 markers below?



A 13 24 14 11 12 14 12 12 11 13 13 29 16 9 10 11 11 25 15 19 28 14 16 17 17
11 11 19 23 15 14 18 17 35 38 12



B 13 24 14 10 12 14 12 12 11 13 13 29 16 9 10 11 11 25 15 19 28 14 16 16 17

11 11 19 23 15 14 18 17 35 39 12



C 13 24 14 11 12 14 12 12 11 13 13 29 16 9 10 11 11 25 15 19 28 14 16 16 17

11 11 19 23 15 14 18 17 35 39 12



D 13 24 14 11 12 14 12 12 11 13 13 29 16 9 10 11 11 25 15 19 29 14 16 16 17

11 11 19 23 15 14 14 17 35 39 12



E 13 24 14 11 12 14 12 12 11 13 13 29 16 9 10 11 11 25 15 19 28 14 17 17 17

11 11 19 23 15 14 19 17 35 39 12



F 13 24 14 11 12 14 12 12 11 13 13 29 16 9 10 11 11 25 15 19 28 14 16 17 17

11 11 19 23 15 14 19 17 35 39 12



G 13 24 14 11 12 14 12 12 11 13 13 29 16 9 10 11 11 25 15 19 28 14 16 17 17

11 11 19 23 15 14 18 17 34 39 12



H 13 24 14 11 12 14 12 12 11 13 13 29 17 9 10 11 11 25 15 19 28 14 16 17 17

11 11 19 23 15 14 18 17 35 39 12



I 13 24 14 11 12 14 12 12 11 13 13 29 16 9 10 11 11 25 15 19 28 14 16 17 17

11 11 19 23 15 14 18 17 36 39 12



J 13 24 14 11 12 14 12 12 11 13 13 29 16 9 10 11 11 25 15 19 29 14 15 17 17

11 11 19 23 15 14 18 17 35 37 12



K 13 24 14 11 12 14 12 12 11 13 13 29 16 9 10 11 11 26 15 19 29 14 15 17 17

11 11 19 23 15 14 18 17 35 37 12



For this group, I have many more haplotypes, but only the above complete to
37 markers. The criteria I used for selecting this group, was that they
were all I had which were complete to 37 markers, so I have not tried to
make some tricky selection to mislead you. Some of the above have been
tested to 130+ markers, and a number to 67 or 76 or more markers.



I don’t know when the common ancestor was for this group, but I know dates
it must have been before, and can speculate dates it was probably before. I
have some speculative ideas about when the common ancestor might have been,
based in part on genealogical data which might imply opportunities for a
common ancestor to have lived.



I don’t know the ancestral haplotype. What do you think the ancestral
haplotype was? How far back do you estimate to the common ancestor using
your system, and what confidence interval range would you give for that
estimate? I have been puzzling over this group for years, and it would be
interesting to see how your estimates compare to my current thinking on the
group. I have an advantage over you in that I do have more marker
information, and some paper trail and context information. But I would be
interesting to see how you evaluate this group based on the limited
information I have given you above.



John.

















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