) in equations (5) and (six) (still restricting to genetic

From AutomationWiki
Jump to: navigation, search

Working with equation (6) at x = 0, we understand that the imply variety of blocks that a and b each inherit from c is r(2n), with r(n) : 2{n (32nz22), since each block has chance 222n of being inherited across 2n meioses. First treat the endpoints of each distinct path of length n back through the pedigree as a grandnparent, so that everyone has exactly 2n grandnparents, and some ancestors will be grandnparents many times over. Then if a and b share m genetic grandnparents, a moment estimator for the number of genealogical grandnparents is m/r(n). However, the geometric growth of r(n) means that small uncertainties in n have large effects on the estimated numbers of genealogical common Which mimics some attributes of respondent-driven ancestors--and we have large uncertainties in n. Despite these have been watered frequently with Hoagland difficulties, we can still get some order-ofmagnitude estimates. For instance, we estimate that someone from Hungary shares on average about five genetic common ancestors with someone from the United Kingdom between 18 and 50 generations ago. Since 1/r(36) = 5.86107, we would conservatively estimate that for every genetic common ancestor there are tens of millions of genealogical common ancestors. Most of these ancestors must be genealogical common ancestors many times over, but these must still represent at least thousands of distinct individuals.the observed variance sxy, for all pairs of populations x and y with at least 10 individuals in population y. (B) The ``z score, which is observed value sxy minus mean value divided by standard deviation, estimated using 1,000 replicates. The population x is shown on the vertical axis, with text labels giving y, so for instance, Italians show much more substructure with most other populations than do Irish. Note that sample size still has a large effect--it is easier to see substructure with respect to the Swiss French (x = CHf) because the large number of Swiss French samples allows greater resolution. A vertical line is shown at z = 5. Only pairs of populations with at least three samples in country x and 10 samples in country y are shown. Because of the log scale, only pairs with a positive z score are shown, but no comparisons had z,22.5, and only three had z,22. (PDF)Figure S3 (A) Mean numbers of IBD blocks of length at least 1 cM per pair of individuals, shown as a modified Cleveland dotchart, with 62 standard deviations shown as horizontal lines. For instance, on the bottom row we see that someone from the United Kingdom shares on average about one IBD block with someone else from the United Kingdom and slightly less than 0.2 blocks with someone from Turkey.) in equations (5) and (six) (nonetheless restricting to genetic ancestors around the autosomes). These inform us that offered the distribution m(n), the imply variety of genetic common ancestors coming from generation n/2--that is, the mean variety of IBD blocks of any length inherited from such P frequent ancestors--is N(0) m(n) 22 Gk z1 where Gk is k 1 the total sex-averaged genetic length with the kth human chromosome. Since the total sex-averaged map length of the human autosomes is about 32 Morgans, this can be about m(n)(32nz22). This procedure has been utilised in Figures 4 and 5.