Final answer:
The probability of producing a child with a dominant phenotype at all four loci in the given cross, with alleles showing incomplete dominance and the genes being unlinked, is calculated using the sum rule and the product rule, resulting in a probability of 81/256.
Step-by-step explanation:
The question asks for the probability of producing a child with a dominant phenotype at all four loci (A, B, C, and D) given that alleles show incomplete dominance and the genes are unlinked. We can find the probability of a dominant phenotype for a single gene by applying the sum rule: the probability of being homozygous dominant (A1A1) is 1/4, and the probability of being heterozygous (A1A2) is 1/2. Therefore, the combined probability is 3/4 for a dominant phenotype at locus A (1/4 + 1/2 = 3/4).
We then apply the same calculations to the B, C, and D loci. Each one has the same probabilities as A because the alleles show incomplete dominance and are unlinked, allowing us to consider each independently. To find the probability of having a dominant phenotype at all four loci together, we apply the product rule: we multiply the individual probabilities of having a dominant phenotype at each locus, which are all 3/4. Thus, the combined probability is (3/4) × (3/4) × (3/4) × (3/4), which equals 81/256.
If we need to verify the combination of probabilities or want a more visual approach, we can use the forked-line method to illustrate the segregation of alleles for each gene and calculate the probabilities for the phenotypes of the offspring from the cross. This method, too, will show that the probability of obtaining an offspring with dominant phenotypes at all four loci is 81/256.