Only One Comparison Needed to Add a Segment to a TG

A Segment-ology TIDBIT

Start with a list of your overlapping shared segments – they all match you (two triangle legs). Find two of these shared segments that match each other (third triangle leg) to form a Triangulated Group! All the rest of the shared segments in your overlapping segment list need only match one of the shared segments in the TG – any one of them with a good overlap – to be added to the TG. The explanation will take a long blog post with diagrams – but the thrust is that forming a TG basically identifies that segment just as good as trio-phasing. So trust me! If a shared segment doesn’t match the TG, it will match the other, overlapping TG; or it is IBC (it happens for some shared segments under 15cM). As a Quality Control measure, I often make a second comparison – it always matches!

 

[22D] Segment-ology: Only One Comparison Needed to Add a Segment to a TG TIDBIT by Jim Bartlett 20170101

11 thoughts on “Only One Comparison Needed to Add a Segment to a TG

  1. I read your work with great interest.
    Perhaps I am missing something – but as I understand it, in the case of a double crossover, two short(er) segments are created – one being maternal and on being paternal – both being the same size.
    If subsequent mitosis events occur in subsequent generations and no crossovers occur to disturb the created segments, many people could have the same segments, half of them from the original maternal parent and half from the original paternal parent. And they would be at the same place on the particular chromosome. So is there not a 50% chance that a segment of that size at that location would be (relative to that ancestor) maternal in origin and would not match those which were paternal? I realize that several conditions must be satisfied for this to happen, but can we really say that a segment as described comes from only one ancestor when there are other segments which came from another ancestor (a spouse?).
    Thank you,
    Vance Wiley

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    • Vance,

      There are several extremely unlikely scenarios that could complicate the issue. There is a chance that segments that are separated could be stitched back together, for instance. In general, each child gets a different, random jigsaw puzzle of segment pieces from independently created sperm and ova. And even if they did have the same, or nearly same, ancestral segment, they would have to pass it down, intact, to descendants who, as cousins, eventually marry. In this case one is a male and one is a female – so it would be possible for a child to get the same ancestral segment on both the maternal and paternal chromosomes (again very unlikely). This would exist for this one generation only. In the next generation, the child can only pass one of those segments on. So in succeeding generations, there would again only be the one ancestral segment. So, at a very remote possibility, two people could pass the same ancestral segment to their child (one on each chromosome), but the child can not pass them both on. You might want to review my blog posts on endogamy. Another way to look at this issue is to understand that you only get paternal segments from your father. He got them from his mother OR father (he can only pass down one or the other). Whoever it was, likewise got that DNA from their mother OR father. Repeat going back as far as you want. Each segment of DNA came from one parent, one grandparent, one Great grandparent, etc. Again, read over my endogamy posts. Jim

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      • Vance, I should add that a double crossover does not create a duplicate segment in a child. There is still only one resulting chromosome that is passed from the parent to the child. The other parent is not involved in this double crossover.

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  2. I have 5 individuals who align up on the same segment area on the same chromosome number (say 8) in GEDmatch chromosome segment matching. However, using GEDmatch one-to-one compare, individual D does not match A or E at that segment area. So that gives me triangulation groups ABC, ABE, ACE, BCD, and BCE. Does that mean that individuals A, B, C, and D match on one of the chromosomes 8 as one big triangulation group and individuals B, C, and D match as one triangulation group on the other chromosome 8?

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