Experimental Hybridizing 91 



some of the descendants, if these albinos are crossed with colored 

 types. Thus if an albino AG be crossed with an albino AB, 

 the albino offspring will be AGAB. Its germ-cells will separate 

 into AG and AB, but these are albinos. If, on the other hand, 

 a black mouse, CB, be mated with an albino A Y, containing 

 yellow as a latent character, the offspring will be CBA Y (yel- 

 low), whose germ-cells will be of four kinds, CB, AY,CY, and 

 AB. Crossing this yellow mouse (with its four kinds of germ- 

 cells) with a white mouse, AGAB, obtained in the way just 

 described, eight possible combinations may follow. The whole 

 process is indicated in the following table : 



Parents AG (albino) AB (albino) 



\ / 



ist generation AGAB (albino) 



AGCB gray (one) 

 ABCB black (one) 

 AGAY] 



2 d generation AGAB ,,. ({ } 



ABAY 

 ABAB] 



yellow (two) 



Cuenot performed this experiment and obtained in the sec- 

 ond generation 151 young, distributed as follows according to 

 color : 



8 1 albinos, 34 yellow, 20 black, 16 gray. 



The probability according to the formula 4 n + 2 n + n+n is: 



76 albinos, 38 yellow, 19 black, 19 gray. 



The agreement is so close that there can be little doubt that the 

 hypothesis is substantially correct. 



Heredity of Piebald or Spoiled Varieties. The piebald condi- 

 tion is regarded by Cuenot as a special mutation, and not one 

 due to crossing colored and white forms. The piebald charac- 

 ter appears in crossing to be dominated by the uniform colora- 

 tion, whatever may be its tint. For example, a spotted gray- 

 and-white mouse crossed by a uniformly black mouse gives a 

 uniformly colored gray mouse, showing that the coupled 



