Final answer:
In a trihybrid cross of AaBbCb×AaBbCc, we use the product rule to calculate the probability of offspring having three dominant alleles. The probability is calculated as (3/4 for A) × (3/4 for B) × (2/4 for C), resulting in 9/64. This demonstrates the principles of independent assortment and dominance in genetics.
Step-by-step explanation:
The cross in question involves three gene pairs, each with a dominant and recessive allele. Specifically, we are examining the F1 cross of AaBbCb×AaBbCc, where 'A' and 'B' are dominant alleles and 'a' and 'b' are their corresponding recessive alleles. The case of 'C' is slightly different, as there is gene interaction where 'C' and 'c' both show a simple dominance relationship. We are interested in calculating the number of offspring that would have 3 uppercase (dominant) alleles. To determine this, we can use either the Punnett square method, the product rule, or the forked-line diagram.
To analyze the cross using the Punnett square is cumbersome due to the large number of genotypes it generates, so the forked-line diagram or product rule is more efficient. Each gene pair is evaluated for its probability of producing a dominant phenotype. For genes A and B, there is a 3 in 4 chance of expressing the dominant phenotype (AA, Aa, or aA; and BB, Bb, or bB). For gene C, it's a 2 in 4 chance because we are combining the heterozygote with a different allele in the partner (Cb, Cc, cc, or cC). We calculate the probability of the offspring having all three dominant phenotypes by multiplying these probabilities: (3/4 for A) × (3/4 for B) × (2/4 for C), which equals 9/64. This means that out of 64 possible offspring combinations, 9 would be expected to have all three uppercase alleles.
This is a classic case of a trihybrid cross demonstrating independent assortment and the principle of dominance. These genetic principles are key to predicting offspring phenotypes from their parental genotypes.