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Subject: Re: [DNA] What shall R1b1c call themselves now?
Date: Wed, 17 Sep 2008 13:26:39 +0000

>From Dennis Wright:
>Can you do the same with any cluster's modals?
For instance, the Atlantic Modal Haplotype AMH at 25 markers is:-
AMH 13 24 14 11 11 14 12 12 12 13 13 29 17 9 10 11 11 25 15 19 29 15 15 17 17

>What you calculate the number of generations for the following Irish clusters since they split from AMH?
A 13 25 14 11 11 13 12 12 12 13 14 29 17 9 10 11 11 25 15 18 30 15 16 16 17
B 13 24 14 10 11 15 12 12 11 13 13 29 17 9 10 11 11 24 15 19 29 15 15 17 17
C 13 24 14 11 11 14 12 12 11 13 13 29 17 8 9 11 11 25 15 19 29 13 13 15 17
D 13 24 14 10 12 15 13 12 12 13 13 29 17 9 10 11 11 25 15 19 29 15 15 16 18

Yes, Dennis, it is principally the same thing as to calculate a time span to a common ancestor. The basic premise is the same, namely, that is takes time for a chain of mutating haplotypes to reach a bottleneck and start their life all over again, as as new ancestral (base) haplotype. The same thing happens with another branch, with another chain of mutating haplotypes, which, having started from the same common ancestor, ends up at a quite another bottleneck, with a different set of mutations (shaped in a presumably statistical way).

Hence, we have two base ("secondary ancestral") haplotypes, which are distant from THEIR COMMON ancestor by X and Y generations, respectively, and each of them is distant from "us" by A and B generations, respectively. Those A and B we determine as "MRCA" for each of them separately.

When you compare their "ancestral" ("secondary ancestral") haplotypes, as shown above, we see a diffence in mutations in each pair. This difference corresponds to X+Y = Z. Therefore, a distance between "us" and the "primary" (in this case) common ancestor is (A+B+Z)/2. It is easy to see if you draw those three nodes on a sheet of paper, that is one "primary" and two "secondary" common ancestors, and a level of "our time", with distances X, Y, A and B.

That I was trying to convey to John, namely that A and B are subjects of familial studies, but all A, B, X and Y are subjects of history-related studies. In a simplest way, of course.

Now, back to your question. It is relatively simple to deal with large differences in mutations (alleles), such as I have shown in those Asian base haplotypes, or between the Asian and European base haplotypes. Clearly, that 8-12 mutations per 12-marker haplotypes, or 23-25 mutations per 25-marker haplotypes show a huge gap between the mutated base haplotypes, and refer to a VERY ANCIENT "primary" common ancestor (which might have happened to be a secondary too, by the way). That is how I came up with 16,000 years for an "age" of R1b haplogroup. I believe that is the primary one (or close to it), since there is no room to go much deeper for it. I have calculated an "age" for R1 in the same manner, and it is around 22,000-25,000 years old (I have to look it up, don't remember exactly now).

However, to calculate the difference between close base haplotypes you have to go to decimals (an average number of their alleles), since we often round them up, and have a difference of "1" when in fact there was "0.2" or even less (10.53 vs. 10.46, for instance become 11 and 10).

In your case a difference between the AMH and the other four base haplotypes (I presume they are not individual haplotypes, but "ancestral"/base haplotypes, aren't they?) is 7, 4, 9 and 6 mutations for A, B, C and D, respectively. I can give VERY rough estimates, but they will be seriously off if you consider decimals in those "round up" differences.

The "gap" between these four pairs (A from the AMH, B from the AMH, etc.) is 184, 98, 249 and 154 generations (with 25 years per generation), that is 4600, 2450, 6225 and 3850 years, respectively. Again, if one considers decimals in their mutational differences, numbers of generations might be noticeable higher or lower. I am just illustrating the principle of calculations. By he way, a correction for back mutation was introduced in each of those four numbers.

Now, if we assume that the AMH is 4000 years (160 generations) "old" and each of those Irish base haplotypes (A-D) is "young", say, 1000 years (40 generations) old, then THEIR (with AMH) common ancestor lived: A/AMH - 4800 years BP, B is derived from the AMH directly, C/AMH - 6400 years BP, D/AMH - 4400 years BP.

To me, the "C" haplotype is particularly interesting. If a difference between C and the AMH is indeed 9 mutations, their common ancestor lived 5,000 years ago even if "C" is very young. If the "age" of "C" is only 40 generations, their common ancestor lived 6400 years ago. If the "age" of "C" is 200 generations, their (with AMH) common ancestor lived 7600 years ago. In this case the haplotype was originated (likely) not in Europe, but elsewhere.

In order to crack that puzzle, we need to know (a) TMCA for "D", and (b) more precise mutational difference between "D" and "AMH" down to 0.1 mutation.

Regards,

Anatole Klyosov