The formula for detemining the number of phenotypic or genotypic classes was: a n Where a is the number of classes (elements, things) observed in the F1 of the cross. And n is the number of such crosses in the multiple gene (trait) cross. 1) Look at the cross A/a; B/b; C/C x A/a; B/b; C/C. We are observing three genes associated with three traits. However, there is variation in to out of the three genes, so in essence this is a dihybrid cross. Lets calculate the phenotypic and genotypic classes: If we look at the monoybrid cross A/a x A/a (which is a part of the dihybrid cross above) we will get F1 with two phenotypic classes Dominant : recessive is 3:1 (D/D+2xD/d : d/d), and three genotypic classes D/D: 2xD/d : d/d is 1 : 2 : 1 So there are two classes (elements, things) when we look at the phenotypes and three classes (elements, things) when we look at the genotypes. Therefore the a in the formula above for phenotypic classes will have the value 2 and for the genotypic classes will have the
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