86 E. LINDHARD 
zygous X awned homozygous Speltoid gives awned homozygotes. 
When U = awnless, u =awned the zygote production by self fertili- 
zation in the awnless Speltoid heterozygote can be thus visualised: 
2 (LUN +n uSp) X S(UN) — 1 UUNN : n uSpUN. 
5. The linkage uSp is so close, that in the progeny of awnless 
heterozygotes of the first 21340 plants only 4 were awned (Tab. 1), 
and as the seed plants were not protected against foreign pollen even 
these might owe their awns to accidental cross fertilization. But in 
Uu-Compactum-heterozygotes this linkage seems partly or entirely 
broken, not converted into repulsion. Therefore it can not be a Mor- 
ganian linkage, u and Sp are not located in the same chromosome, the 
conception being that in the Speltoid-heterozygote the two chromoso- 
mes carrying u and Sp constantly follow each other by the building of © 
gametes and are both eliminated together in the building of 4 gametes, 
the linkage thus forming a kind of heterogamous chromosome binding. 
6. The sex differentiation following, and possibly causing the chro- 
mosome-binding, is termed heterogamy (H. pe Vries, NizssoN-EuLE). In 
the Speltoid-heterozygote heterogamy is complete when the reduplica- 
tion of 2 uSp gametes and the elimination of d uSp gametes are 
complete so that only uSpUN zygotes are produced, and it is partial 
when either the reduplication or the elimination is partial. 
7. Partial heterogamy as shown through the ratio of segregation 
may remain stable through several generations, but the modus of se- 
gregation may suddenly change, (Umschlag, Nitsson-Eute). By such 
reversions from complete to partial elimination of uSp gametes, homo- 
zygous Speltoids can segregate from lines hitherto yielding Normal 
type and heterozygotes only. 
8. The ratios of segregation for lines spontaneously segregating 
Speltoid-homozygotes and for the F, from artificial crosses of Speltoids 
can be grouped under one of the following headings: 
1) In two cases we obtained: 
No elimination of 4 Sp gametes reduplication 8 of 2 Sp gametes or 
@ (1N+8 Sp) X g (N+ Sp) — 1 Normal: 9 heteroz. : 8 homoz. 
2) In 6 cases: 
elimination ‘4, no reduplication, thus 
SN + Sp) XS (8 N +1 Sp) =8 Norm.:9 het.: 1 hom. 
3) In 6 cases: 
elimination */,, reduplication 8, giving 
2(1 N+8 Sp) X (4 N +1 Sp) =4 Norm.: 33 het.:8 hom. 
