No. 562] THE PROBLEM OF INBREEDING 



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the sixth generation, and so on until in the twelfth genera- 

 tion 256 ancestors of Hamlet will be so eliminated. The 

 same consideration applies in every other like case. 



Practically then the method of dealing with a pedigree 

 of this sort is first to go through and indicate in a distinc- 

 tive way every primary reappearance of individuals. 

 Then form a table on the plan of Table IV, the character 

 of which is so obvious as not to need detailed explanation. 



This table is to be read in the following way : Because 

 of the reappearance of Hamlet in the fourth ancestral 

 generation Saxton has 1 fewer ancestors in that genera- 

 tion than he would have had in the entire absence of in- 

 breeding; 2 fewer in the fifth generation and so on. The 

 totals of the columns of this table are the values, for 

 each generation, of 



p n+1 — q n+1 



in (iii). These totals, multiplied by 100, have then merely 

 to be divided by p n+1 in order to obtain the successive Z's. 

 The whole operation may be very quickly carried out. It 

 is not in fact necessary to fill out the whole of the later 

 columns of the table, the entries may be cumulated. 

 For the present pedigree we have 



Z =0, as alwavs, 10 



Z, =0, 



Z 2 = 0, 



Z 9 = 6.25 per cent., 

 Z 4 =400/32 = 12.50, 

 Z B =1,100/64=17.19, 

 Z 6 =2,700/128 = 21.09, 



