12 (204) SURPASSING UPWARD AND DOWNWARD OF THE INDICES 
So in this case there will be a surpassing of the indices of parents by 
those of children 1). 
In this way on the base of segregating heredity the evidently pres- 
ent phenomenon that there are more families in which the indices of 
children surpass those of parents upward than downward has been ex- 
plained. 
The result of the admitted segregation that the tables IIa and Illa 
show may have been obscured by non-hereditary variability. Non- 
hereditary variability must influence the result in the direction of too 
many heterozygotes. In grouping the indices of children which agree 
with the lowest index (table II), respectively the highest index (table 
III) of parents are brought into one group and all other different high 
ones into a 2d group. It is evident that if there is a large non-heredit- 
ary variability, the 2nd group must contain indices which properly 
belong to the 18t group. 
That for table IIa we find an other proportion (1 : 5) than for table 
Ila (1 : 2.75) can be explained as follows. In table II the low indices 
form the group of the non-varying individuals, i. e. of the homozygotes 
(RR). Accepting some dominance of the high indices over the low ones, 
then, of the heterozygotes which form the group of different high in- 
dices of table II, there will be brought but few — being phaenotypicallow 
indices — into the group of the low indices. In table IIa on the contrary 
in which the group of the high indices forms the homozygotes DD, a 
great deal of the heterozygotes — again, the high indices being more or 
less dominant — will be brought into the group of the high indices. 
Thus in table IIIa the proportion of the homozygotes and the heterozy- 
gotes becomes larger than in table IIa. 
So the tables Ila and IIIa do not show that the children of these fam- 
ilies form 2 about equal groups according to the formulae aa x Aa — 
aa + Aa, or Aa x Aa = AA + Aa (+ aa) and AA x Aa=AA=+ 
Aa Or Aa x Aa (AA) + 2Aä aa: 
It is possible that the result is counterbalanced by non-hereditary 
1) The reverse, the P-forms, as we bespoke for the explanation of the cases of tab. 
II, being 80 and 75 according to the formula DR x RR and the parent with the index 
80 being a dominant heterozygote of indices 80 and 70, will occur also, but then 
among the children indices smaller than 75 will be found extremely rarely, just as there 
is dominance of the high index. So the acceptation for tab. II does not procure cases 
for tab. III, those of tab. III foı tab. II, as we saw, they do, 
