LA. THEORIE DU CROISEMENT. 
209 
„Theii if 71 = the nuinber of successive self-fertilisations 
„and m = tlie nuinber of pairs of characters, 
^ = 1 — (y + 
2»m — (2» 
1)' 
2' 
(1) 
(2) 
(3) 
(4). 
„Examples: (1). Suppose tliat there hâve beeu eiglit self-fertilisations 
,,and that we are dealing with 10 pairs of characters. What proportion 
„x of the organisms will be homozjgotic with respect to ail the 10 char- 
„acters? What proportion will be homozygotic with respect to anj given 
,,one character? To any given two or three? 
jjTaking first the case for the entire characters by formula (1) 
(2,8 1 V 10 /255\ 
-^^J = = log 9,9830020 = 0,961617. 
,,Thus out of 100 individuals somewhat above 96 would be pure 
^homozygotes, or by formula (4) but one in 26 would be heterozygotic 
„in any respect [v = 0,038383). 
„With respect to any one character formula (1) gives: 
0,99609375 
„so that ail but 4 in 1000 would be homozygotes with respect to that 
„ character. 
„In the same way we find that with respect to any two characters the 
„proportion of homozygotes would be 0,9922, with respect to three 
,,0,9883, with respect to four 0,9845 etc. 
„(2) Suppose that there are 20 pairs of characters and that there havC 
„been 20 self-fertilisations. Tlien: 
/^^^ 1\^^ /l 048575\^^ 
^ = =(i:0485r6) =% = 0,99998. 
ARCHIVES NÉERLANDAISES, SERIE III B, TOME II. 14 
