RECOMBINATION IN SEXUAL ORGANISMS 89 



larger numbers of independently segregating genes are involved in a 

 cross. 



Because of the problem of dominance, we are concerned with the 

 phenotypes of the offspring as well as the genotypes. A short-cut 

 method may be used. Considering one factor at a time, we know that 

 the recessive phenotype will occur with a frequency of \. Then: 



/4 X /4 = Vie blue-eyed, galactose nonutilizers 

 /^ X ^ = ^16 blue-eyed, galactose utilizers 

 % X y^ = "/^le brown-eyed, galactose nonutilizers 

 '/4 X % = ^ig brown-eyed, galactose utilizers 



Next let us see how linkage and crossing over affect the transmission 

 of genes in diploid organisms. Suppose the genes for blue eyes and 

 galactose utilization in man were linked and not, as was presumed in the 

 case illustrated in Figure 3.11, on different chromosomes. Then, if 

 crossing over occurred in 50 per cent of the meioses, the haploid gametes 

 of individuals heterozygous for these genes (AB/ab) would not occur 

 with the genotypes AB, ab, Ab, and aB in equal numbers. On the con- 

 trary, their frequencies would be 0.375, 0.375, 0.125, and 0.125 respec- 

 tively (0.5 X 0.5 = 0.25, which is the frequency of old and of new combi- 

 nations produced in meioses in which crossing over occurs; the frequency 

 of each type of new and old combination is, therefore, 0.125). The last 

 two classes would be the new combinations. Matings of individuals 

 heterozygous for these genes would give a complicated ratio of types 

 among their offspring. Multiplying the expression 0.375AB -|-0.375afe-|- 

 0.125A/? -(-0.125aB by itself would give a ratio of brown-eyed utilizers: 

 brown-eyed nonutilizers : blue-eyed utilizers : blue-eyed nonutilizers of 

 about 64:11:11:14. This might be distinguished from the 9:3:3:1 

 ratio which is achieved when the genes are unlinked. But the recom- 

 binations can be more simply detected if the heterozygote is crossed to a 

 recessive homozygote whose gametes are all ab in composition, a so- 

 called test cross. In this case, 0.375 of the offspring would be brown- 

 eyed utilizers (AB/ab), 0.375 would be blue-eyed nonutilizers (ab/ab), 

 0.125 would be blue-eyed utihzers (Ab/ab), and 0.125 would be brown- 

 eyed nonutilizers (aB/ab), and the frequencies of recombinant gametes 

 from the heterozygote could be determined directly. 



Not many human families are large enough to realize even simple 

 ratios which are only the averages found in ideal populations of infinite 

 size. Man is an inconvenient object for genetic study, although we want 

 more to know about his genetics than any other. He reproduces slowly 

 and has small families so that the offspring of many parents of identical 

 genotype must be counted to get a large enough set of data in which to 



