68 



PHYSICAL BASIS OF HEREDITY 



Eggs 

 BW 



Bw 



bW 



bw 



Sperm 

 BW 



Bw 



bW 



bw 



The resulting ratio is 9 grays, 3 blacks, and 4 whites. The 

 last two terms of the 9:3:3:1 ratio are here united in one 

 class (4 whites) because when homozygous for absence of 

 color the individual is white, regardless as to whether 

 the other color-producing factors make for the wild type 

 of coloration or for some mutant color. 



Another interesting two-pair case involves varieties 

 of the combs of domesticated breeds of fowls. There is a 

 dominant type called "Rose" (Fig. 31, c), which, bred to 

 single (wild type. Fig. 31, a), gives Rose in F^, and 3 Rose 

 to 1 Single in F2. Another dominant type called "Pea" 

 (Fig. 31, b) likewise gives Piea in F^ and 3 Peas to 1 Single 

 Comb in F2. But when Rose is bred to Pea there is not 

 produced the wild type, as one might have anticipated, but 

 a comb called "Walnut" (Fig. Sl,d), that differs from 

 both parental types. The character is due to the com- 

 bined action of both dominants. If two F^ birds with 

 Walnut combs are bred to each other they give 9 Walnut, 

 3 Pea., 3 Rose, 1 Single comb. This ratio shows that two 

 factors are involved, and that the Walnut comb appears 

 in all birds carrying both the Rose and the Pea genes. 

 The Single comb is the double recessive form. 



