THE PROBLEM OF INBREEDING 153 



+ 16 Si (48) + i (80)J = 16 X 44 = 704 (AA + 



Aa) families 



+ 16 Si (80) I = 160 Aa families 

 + 16 Si (48) + 8 + J (80)| = 16 X 52 = 832 (AA 



+ 2 A a + aa) families 

 -f 704 (Aa + aa) families 

 + 848 aa families. 



Succeeding generations follow the same law and 

 need not be worked out in detail. 



So far the discussion has confined itself to 

 families, as this must be the basic unit in the 

 theory of any form of inbreeding. Turning to 

 individuals, we have the following simple relations 

 to pass to individuals : 



In the nth generation the number of 

 A A (or aa) individuals = 2 s (o n ) + s (p n ) + 



i S (Tn). 



Aa (or a A) individuals = 2 s (q n ) + s 



The first of the above expressions multiplied 

 by 2 gives the total heterozygotes. 



The results under certain conditions of brother 

 X sister and cousin mating have been discussed by 

 Jacobs. 1 



Jennings 2 in a very interesting and valuable 



1 Jacobs, S. M. "Inbreeding in a Stable Simple Mendelian 

 Population with Special Reference to Cousin Marriage. Proc. Roy. 

 Soc., Vol. 84, B, pp. 23-41, 1911. 



2 Jennings, H. S. "Formulae for the Results of Inbreeding." 

 Amer.,Nat., Vol. XLIII, pp. 693-696, 1914. 



