The number of expected genotypes to compare the observation to would be 20 RrSs, 20 Rrss, 20 rrSs and 20 rrss
Dihybrid crossesCrossing RrSs with rrss will yield offspring with the following phenotype according to the attached Punnet's square:
4RrSs
4Rrss
4rrSs
4rrss
Total offspring = 7+35+32+6 =
Thus instead of 7:35:32:6, it will be 1:1:1:1 if the two loci were independently assorting. The number of expected genotypes would be 20 RrSs, 20 Rrss, 20 rrSs and 20 rrss
More on independent assortment can be found here: