60 HEREDITY 



four, and shade three-fourths of each in the opposite 

 direction to represent the fact that three-fourths are 

 hornless, one-fourth horned. We have now a figure 

 divided into sixteenths. Of these subdivisions, nine 

 are shaded in both directions, representing nine in- 

 dividuals containing both dominants i.e. black and 

 hornless. Three squares are shaded in one direction 

 only, representing three red and hornless individuals. 

 Three are shaded in the other direction only i.e. three 

 individuals will be black and horned. One square 

 only remains unshaded on the average only one 

 individual in sixteen will be red and horned, and will 

 show both recessives. 



To state the matter differently, out of sixteen 

 individuals twelve or three-fourths will be black, and 

 four or one-fourth red. Of the twelve blacks, nine or 

 three-fourths will be hornless, and three or one-fourth 

 horned. Of the four reds three or three-fourths will 

 be hornless, and one or one-fourth will be horned. We 

 therefore have the ratio V 



9 with both dominants. 

 3 with one dominant only. 

 3 with the other dominant only. 

 1 with both recessives. 



Very close approximations to this 9:3:3:1 ratio 

 have been obtained in a large number of instances. 



Of the more complicated instances of Mendelian 

 inheritance, one of the best known and most interesting 

 is that of the type of comb in fowls. The original form 

 of comb, that found in the wild jungle fowl, is the 

 " single " (C in Fig. 9). Other forms are the " rose " (B), 

 "pea" (A), and " walnut" (D). The rose behaves as a 

 simple dominant to the single. The first hybrid genera- 

 tion have rose combs, and of the second 75 per cent, are 

 rose, 25 per cent, single. Pea is also a simple dominant 

 to single, giving the ration 3 pea : 1 single in the second 

 generation. Walnut behaves as a simple dominant to 

 both pea and rose. 



If walnut be crossed with single the first generation 



