128 



Dihybrids, Trihyhrids, and Polyhybrids 



Since dominance of both genes is complete, the phenotype ratio 

 in the F2 is 9 Dominant Dominant : 3 Dominant recessive : re- 

 cessive Dominant : 1 recessive recessive. 



The checkerboard is an easy means of showing the results of 

 combining the four eggs with the four male gametes at random. 

 It is somewhat cumbersome, however, and need not be used if 

 the student approaches the F2 as a problem of combined ratios. 

 If for the moment only the PI and pi genes are considered, the 

 Fo genotypic ratio is 1 PlPl : 2 Plpl : 1 plpl. If, the genes C7' 



Male Gametes 

 PlCr Pier plCr 



pi cr 



PlCr 



m PI cr 



bo 



pl cr 



Fig. 41, Checkerboard method of determining the F2 from a cross be- 

 tween a homozygous purple, normal-leaved maize plant of the genotype 

 PlPl CrCr and a green, crinkly-leaved plant, plpl crcr. 



and cr are considered, the F2 genotypes are 1 CrCr : 2 Crcr : 1 

 crcr. Now when we combine two independent monohybrid ratios 

 we see that one-fourth the PlPl plants will also be CrCr, that one- 

 half will be Crcr, and that the remaining one-fourth will also be 

 crcr. The same is true of all the Plpl and also of all the plpl 

 plants. The method of obtaining a dihybrid Fo genotypic ratio 

 by combining the two independent F2 monohybrid genotypic 

 ratios of which it is composed is shown in Fig. 42. It is easily 

 seen that the results are the same as those obtained by the 

 checkerboard method. 



In a similar manner, the dihybrid phenotypic ratio can be 

 determined by combining the two monohybrid phenotypic ratios 

 (Fig. 42). Although the mdependent monohybrid ratios have 

 been presented in the form of ratios of whole numbers (3 : 1), 

 they could also be written in the form of ratios of probabilities 

 (% : %). The probability of getting a purple plant is % and 

 the probability of getting a noncrinkly plant is %. Therefore, 

 the probability of getting a plant that is both purple and non- 

 crinkly is % X /4 or % 6- The probabilities of the other com- 



