60 GENETIC STOCKS AND BREEDING METHODS 



2 ;„ _ (86 - 93 + 35 - 15) 2 169 = Q ^ 



XJ e 



471 471 



0T (86 - 93 - 105 + 45) 2 4489 

 % L = I4T3 = T4T3 = 3 - 177 



The £ 2 's for the segregation of ru andje are very small. That for linkage is larger, 

 3.177, but still below the 5 per cent level of significance for one degree of freedom. 

 The data thus do not give evidence for linkage. 



It should be noted that the sum of the three values of y 2 calculated above, 3.555, is 

 equal to the total % 2 calculated by the standard formula. This will always be so if the 

 arithmetic has been correctly performed. 



If the deviations of the single factor segregations from expectation are significant 

 because of reduced viability, reduced penetrance, or any other disturbing factor, the 

 above formula for the linkage % 2 does not give a true estimate of the significance of the 

 deviation due to linkage. It will be better in this case to calculate a contingency % 2 to 

 determine whether the two pairs of genes are recombining at random, regardless of 

 their individual segregation ratios. This can be done by the standard method, calcu- 

 lating the expected frequency in the four classes from the marginal totals. A somewhat 

 simpler method is to use the special formula applicable to two-by-two tables, which 

 eliminates the necessity for calculating the expected frequencies. By this formula, 



2 «( aifl 4 - Q 2 Q 3 ) 2 



X = ~ 



K + «2)( fl 3 + a i )(a l + a 3 )(a 2 + a 4 ) 



where a u a 2 , . . . , have the same meaning as in table 14. This % 2 has one degree of 

 freedom. Its use can be illustrated by the following example. 549 



The segregation of ogligodactyly, ol, and albino, c, in the mouse in an intercross in 

 repulsion was as follows : 



+ + +c ol + olc Total 

 334 188 107 13 642 



The segregation of both ol and c is very abnormal as shown by % 2 {% 2 ol — 13.6, y^c = 

 13.6), there being a deficiency of ol and an excess of c. When the data are arranged in 

 a two-by-two table, it appears that there is an excess in the +c and ol+ classes, as 

 would be expected in the case of linkage. 



+ c Total 



+ 334 188 522 



ol 107 13 120 



Total 441 201 642 



To assess the significance of this excess we calculate % 2 by the above formula. 



642(4342- 201 16) 2 _ 

 z 522 x 120 x 441 x 201 zo - /u ^ 



For one degree of freedom this is highly significant. 



