38 INTROGRESSIVE HYBRIDIZATION 



plasm. Let us take the simple example of a short chromo- 

 some in which there is regularly a single crossover. Let us 

 further suppose that in the 2 species, or races, which are to 

 be crossed, there are 10 pairs of gene differences within this 

 chromosome. This seems a conservative number for a 

 length of germplasm which might well be 50 units long 

 genetically and made up of 200 or more genes. 



In the gametes of the first-generation hybrid, as a result 

 of 4-strand crossing over, one half of the gametes will have 

 one crossed-over section in this chromosome and the other 

 half will have none. The number of crossovers per chromo- 

 some will be increased the same way in each generation: 

 Double crossovers will not be possible until the F2 genera- 

 tion forms its gametes, triple crossovers until the F3, etc. 

 In each generation one half the gametes wall acquire an extra 

 crossover, one half will continue the previous number. The 

 number of crossovers per gamete and the proportions of each 

 kind of gamete can therefore be obtained from expanding 

 {}i + M),'^ in which n equals the number of hybrid gen- 

 erations. For the 10 gene pairs under consideration complete 

 recombination cannot be attained until gametes are pro- 

 duced in which all 9 breaks between the original sets of 10 

 differing gene pairs have occurred. To obtain such a gamete 

 will require a minimum of 9 hybrid generations, and even 

 then these gametes may be expected only once in 2^ (= 512). 

 It will require twice as many hybrid generations before gam- 

 etes of this degree of recombination will be in the majority. 



A more precise estimate of the hindrance to recombina- 

 tion can be obtained by considering the ratio of the possible 

 gene combinations in the germ cells of Fi to random com- 

 bination. With 3 pairs of differing loci, abc/ABC, there can 

 be a crossover between the a locus and the h locus and be- 

 tween the b and the c. Each of these will permit two recom- 

 binations, viz., aBC, Abe, and ahC, ABc. The total number 

 of recombinations will therefore be equal to twice the num- 

 ber of gene abutments or 2(n — 1), in w^hich n equals the 

 number of differing gene pairs. With the two original com- 



