PLANT GENETICS 31 



"3. The variability of the F 2 populations produced from such crosses 

 should be much greater than that of the FI populations, and if a sufficient 

 number of individuals are produced the grand-parental types should be 

 recovered. The fulfillment of this condition comes about from the 

 general laws of segregation of factors in F\ and their recombination in JFV 



"4. In certain cases F 2 individuals should be produced showing a 

 greater or a less extreme development of the character complex than 

 either grandparent. This is merely the result of recombination of 

 modifiers, as was explained above. 



"5. Individuals of different types from the F generation should 

 produce populations differing in type. The idea on which this state- 

 ment is based is, of course, that all F 2 individuals are not alike in their 

 inherited constitution and therefore must breed differently. 



"6. Individuals either of the same or of different types chosen from 

 the F z generation should give F z populations differing in the amount 

 of their variability. This conclusion depends on the fact that some 

 individuals in the F^ generation will be heterozygous for many factors 

 and some heterozygous for only a few factors." 



From the standpoint of the student a hypothetical case may 

 be given to show how the factor hypothesis may be used to 

 explain the inheritance of quantitative characters. Given two 

 barley varieties as follows: 



t Variety 1, average length of internode of rachis 2.0 mm. 

 Variety 2, average length of internode of rachis 3.6 mm. 



Suppose these varieties differ by two separately inherited factors, 

 A and B, each when homozygous causing a lengthening of the 

 internode by 0.8 mm.; when heterozygous by 0.4 mm., 



Variety 1 aabb Gamete ab _, 



XT- A n A A rt-o r* A n *! ZygOte AdBb 



Variety 2 A ABB Gamete AB 

 Combinations in F 2 would occur as follows: 



Fz PLANTS Fa BKEEDING NATURE 



1 AABB Would breed true for length of internode of 3.6 mm. 



2 AaBB Would segregate from 3.6 mm. to 2.8 mm. 

 2 AABb Would segregate from 3.6 mm. to 2.8 mm. 

 4 AaBb Would segregate as F 2 . 



1 AAbb Would breed true for length of internode of 2.8 mm. 



2 Aabb Would segregate from 2.8 to 2.0 mm. 



1 aaBB Would breed true for length of internode of 2.8 mm. 



2 aaBb Would segregate from 2.8 to 2.0 mm. 



1 aabb Would breed true for length of internode of 2.0 mm. 



Probably few size characters are as simple in their inheritance as 

 this illustration. However, the factor notation assists in gaining 



