Final answer:
In a dihybrid cross aabb × aabb, all offspring will be homozygous for the recessive traits, making the fraction 1/4. Additionally, in a tetrahybrid cross of heterozygotes (AaBbCcDd), the probability of offspring with dominant phenotypes at all loci is 81/256, while the chance of being homozygous recessive for all four alleles is 1/256.answer is option A.
Step-by-step explanation:
In the dihybrid cross aabb × aabb, all the offspring will inherit the aa and bb genotypes from both parents, because the parents can only pass on the alleles they possess, which in this case are both homozygous recessive. This means that the fraction of the offspring predicted to be homozygous for both recessive traits is 100%, because there are no other alleles present in the parents to be passed on. Therefore, the correct answer is Option 1: 1/4.
Using the sum and product rules for probabilities, in the case of a tetrahybrid cross where individuals are heterozygous for all four genes (AaBbCcDd), the expected proportion of offspring displaying the dominant phenotype for all four loci would be calculated by adding the probabilities of being homozygous dominant or heterozygous for each gene (3/4 for each), and then multiplying those probabilities together (3/4 × 3/4 × 3/4 × 3/4), resulting in 81/256.
For homozygous recessive outcomes at all four alleles in a tetrahybrid cross, since each allele has a 1/4 chance of being homozygous recessive, the combined probability is (1/4) × (1/4) × (1/4) × (1/4), which equals 1/256.