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09.07.2023, solved by verified expert

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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:

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