When crossing AaBb x AaBb, what fraction of the offspring will be homozygous for both recessive traits?

To find the fraction of offspring that will be homozygous for both recessive traits when crossing AaBb x AaBb, it helps to visualize the genetic combinations that can arise from this dihybrid cross.

Each parent can produce gametes with the following combinations of alleles: AB, Ab, aB, and ab. The resulting Punnett square will include all possible combinations of these gametes, totaling 16 squares, as indicated by the 4 gametes from each parent.

The goal is to find the homozygous recessive genotype for both traits, which in this case corresponds to the genotype "aabb." To determine how frequently this genotype appears in the offspring:

1. Identify the probability of each trait being recessive (since both A and B are dominant):

• The probability of getting "aa" (homozygous recessive for the first trait) from Aa x Aa is 1/4.

• The probability of getting "bb" (homozygous recessive for the second trait) from Bb x Bb is also 1/4.

2. To find the probability of both traits being homozygous recessive simultaneously, multiply the probabilities of each independent event:

• The calculation is