44 BULLETIN 1121, U. S. DEPARTMENT OF AGRICULTURE. 
It is not difficult to find the percentage of heterozygosis under 
random breeding from stock derived from any given number of 
families. Assume that there are n families. Symbolize each family 
by a letter P,Q, R,S, etc. There are 5 (n—1) different first crosses 
(PQ, PR, PS, QR, ete.). On commencing random breeding there is 
one chance in 5 (n—1) of making a mating in which both families 
are used twice (like PQx PQ) and which is thus equivalent to a 
mating of experiment Cl. There are 2 (n—2) chances in ; (n— 1} 
of making matings in which one family only is used twice, (like 
PQxQR). There are $(n—2) (n—3) chances in 5 (n—1) of making 
matings in which neither family is used twice, (like PQ RS) and 
which are equivalent to Experiment CC. In the last case, as we have 
seen, there is complete recovery, on the average, of the heterozygosis 
of the original random-bred stock. In the first case (PQ x PQ) there 
is only half recovery. One might expect to find an exactly intermedi- 
ate result in the case (PQ x QR), and this can easily be shown to be 
true by the use of path coefficients (Wright, 1921). 
Since P, Q, R, and S are assumed to be completely homozygous 
inbred families, the constitution of the germ cells is completely 
determined. Thus the path coefficient from. zygote to germ cell 
(b’) is 1.0 in the first generation. As there is assumed to be no 
correlation between the families and hence none between their germ 
cells (f’=0), the coefficient for the degree of determination of the 
progeny (first cross) by germ cells (a’*) equals 4. For the cor- 
relation between mated individuals of the next generation we have 
in consequence m= 1 in the case PQ x PQ; m=41in the case PQx QR, 
and m=O in the case PQx RS. For the path coefficient, zygote to 
germ cell, second generation, we have b= ./1, by the formula 6?= 
4(1+/’). For the correlation between uniting germ cells of this 
generation we have f=b?m. Finally, by the formula for percentage 
of heterozygosis, p=2zy (1—/f), we have p=zyin the case PQ x PQ, 
p=3/2 xy in the case PQxXQR, and p=2zy in the case PQx RS. 
Thus the second case is exactly intermediate between the others, as 
we set out to prove. 
Multiplying the number of matings of each kind by the correspond- 
ing percentage of heterozygosis and adding, we find the total per- 
centage of heterozygosis in the new random-bred stock to be 
xy (m= 1) 2ry(n— 1) 
100 aoe aie Se 
in the original random-bred stock. Since each pair of factors comes 
to equilibrium after one generation of random mating, this level of 
where 100 x 2zry is the percentage 
