128 THE FIRST PRINCIPLES OF HEREDITY 



recessive characters are indicated by a (green) and b (angular) 

 respectively, then the two pairs of allelomorphs will be Aa 

 and Bb. The hybrid (FJ will have the constitution ABab, 

 but will appear like AB, because A and B are dominant 

 over a and b. 



We can easily find the proportion of the F 2 generation. 

 Taking the two pairs of allelomorphs, the offspring will be : 

 three dominant to one recessive for each pair of allelo- 

 morphs namely (taking sixteen Individuals for Dihybrids), 

 12 A to 4 a and 12 B to 4 &. But now a recombination 

 takes place again in the proportion of three dominant to 

 one recessive. Out of the 12 A, 9 A combine with 9 B 

 (dominant), and 3 A with 3 b (recessive). In the same way, 

 out of the 4 a, 3 a combine with 3 B (dominant), and i a 

 with i b (recessive), so that we get as result : 9 AB + 3 

 A& + 3 aB + 1 ab. How can this result be explained in 

 terms of gametes ? 



As the two pairs of allelomorphs are Aa and Bb, and as, 

 according to the law of gametic segregation, only one 

 unit of each pair of allelomorphs can be contained in the 

 same gamete together with any other unit,, we get as 

 gametes of the hybrid the following four possible combina- 

 tions : AB, Ab, aB, and ab. On further inbreeding of the 

 F! generation, all the possibilities of the offspring can once 

 more be graphically represented by means of squares, 

 according to Punnett's method, in the following manner : 

 We take this time sixteen female gametes and sixteen male 

 gametes, four each of the above given constitutions, and 

 arrange them, the female gametes vertically and the 

 male horizontally. Then we superimpose the squares, 

 and find thus all the possible combinations of the F 2 

 generation. 



We find as result the following combinations : 



1 ABAB^ II. i AbAb\ 



2 ^ h AB. . TTT 2 A** 

 2 ABaB I III. i aB a 



4 ABafc J 2 aBab 



AB. . TTT 2 - ' IV . , abab . 



III. i aB aB 



