Final answer:
The predicted number of individuals with genotype a+a bb c+c from a trihybrid testcross can be calculated using Mendelian genetics principles and the product rule; however, none of the given multiple-choice options match the calculated expected number of 63 individuals when applying these principles to 1000 offspring.
Step-by-step explanation:
The question deals with predicting the genotypic outcome of a trihybrid cross between individuals of specific genotypes. When calculating the number of expected individuals with the genotype a+a bb c+c in a trihybrid testcross, we can apply the principles of Mendelian genetics.
For the a locus, since one parent is aa and the other is Aa, 50 percent of the offspring would be a+a (or aa). For the b locus, we are looking for the homozygous recessive genotype bb, which will occur 25 percent of the time (as it is a testcross with one parent being homozygous recessive and the other heterozygous, resulting in a 1:1 ratio). And for the c locus, the genotype c+c (or cc) will also occur 50 percent of the time for similar reasons.
Therefore, by using the product rule to calculate the combined probability of all three independent events occurring simultaneously (0.5 x 0.25 x 0.5), we get 0.0625 or 6.25 percent. When we apply this percentage to 1000 offspring, we multiply 0.0625 by 1000, yielding 62.5, which rounds to 63 expected individuals with the genotype a+a bb c+c.
Since this specific number is not provided in the multiple-choice options, the correct procedure to calculate the number, rather than the final result, should be discussed.