Aug. 15,1925 
Coefficients of Inbreeding and Relationship 
381 
inbreeding indicated by a two-column pedigree, it is not necessary to 
count the generations to the closest common ancestor; it is merely 
necessary to note whether there is a tie, and what animal is responsible 
for it. Neglecting for the moment the effect of inbred common 
ancestry, the coefficient merely takes the values 50 per cent and 
0 per cent under the two alternatives. As noted before, such a 
determination means practically nothing as far as the individual is 
concerned. But by determining the proportion of such ties in a 
sufficiently large random sample of a family or breed, a measure 
of the average degree of inbreeding of that family or breed can be 
obtained to as high a degree of accuracy as desired. If, for example, 
40 two-column pedigrees show a common ancestor and 60 do not, an 
average inbreeding of 20 per cent ( = 40X0.50) is indicated, again 
neglecting the term (1 + F a ). 
The probable error of the percentage of ties can be calculated by 
v 
pq 
N’ 
where N is the number of cases. 
the usual formula, = 0.6745* 
p is the observed chance of occurrence of a tie, and q (= 1 — p) is the 
chance of nonoccurrence. In the case cited, E p = 0.6745^— 
40X0.60 
100 
= 0.032. As 40 per cent of ties corresponds to an inbreeding co¬ 
efficient of 20 per cent, the probable error of the latter must be rated 
down proportionately, giving 20 ±1.6 per cent. Allowance for the 
factor (1 + F a ) will be considered later. 
The method may be used to calculate the inbreeding coefficient of 
an individual, by finding the percentage of ties in a large number of 
random two-column samples from his pedigree. Increased accuracy 
may be obtained in this case, however, by a combination of the approx¬ 
imate method with the complete method. The pedigree may, for 
example, be tabulated completely for five generations, continuing 
each of the 32 lines to the foundation stock by the random method. 
As there are 16 random lines back of the sire and 16 back of the dam, 
there are 256 tabulated pairs of lines to be considered for possible ties. 
The total number of possible pairs of lines tracing n generations back 
2 n+nl 
of the sire and n 1 back of the dam is thus times the sample, 
if n and n 1 are both greater than four in this case. The contribution 
of an observed tie this distance back in the pedigree is (J^) n+nl+1 
(1 + FJ. Multiplying by the number of pairs of lines which it 
represents gives (1 + F a ) as the contribution to be assigned the 
observed tie as a sample of the complete pedigree. This expression, 
it will be noted, is free from n and n, 1 as in the two-column case. It 
is easy to see that, in general, the contribution to be assigned to a 
tie in the random part of a pedigree which is complete for Tc genera¬ 
tions back of the parents is (}4) 2k+1 (1 + F<j) • 
If a common ancestor appears in the complete portion of the 
pedigree on one side and in the random portion on the other side, 
this formula requires modification. Letting n be less than Tc but 
n 1 greater, the ratio of possible to tabulated pairs of lines tracing n 
generations back of the sire and n 1 generations back of the dam is 
2 n+ri 1 2 n l 2 71+711 . 
2i+F = 2*’ instead of as in the case previously considered. 
