Final answer:
The probability of an offspring having both dominant traits A and B from a cross with a heterozygote AaBb is 9/16. This is calculated by multiplying the individual probabilities of getting a dominant allele from each gene, both of which are 3/4, using the product rule.
Step-by-step explanation:
The question involves using Mendelian genetics to determine the probability of offspring inheriting dominant traits in a dihybrid cross. To calculate this, we will apply both the sum and product rules of probability. Given that one parent is AaBb, we know that the alleles for traits A and B can segregate independently, and the possible combinations in the gametes are AB, Ab, aB, and ab.
For a genotype to show dominant traits, it can either have homozygous dominant (AA or BB) or heterozygous (Aa or Bb) alleles. The probability of obtaining a dominant allele A from Aa is 3/4, since it could be AA (homozygous dominant, with a probability of 1/4) or Aa (heterozygous, with a probability of 1/2). The probability of obtaining a dominant allele B from Bb is also 3/4. Using the product rule to calculate the combined probability of having both dominant traits, we multiply the individual probabilities: 3/4 (for A) × 3/4 (for B). The result is 9/16.
Hence, the probability of an offspring having both dominant traits from a cross with AaBb is 9/16 or approximately 56.25%..