Final answer:
An individual with aB/Ab genotype will produce approximately 85% parental gametes (aB and Ab) and 15% recombinant gametes (AB and ab) due to crossovers between the linked genes A and B that are 15 cM apart.
Step-by-step explanation:
The gametes that are produced by an individual with the genotype aB/Ab when the genes (A and B) are on the same chromosome and the distance between them is 15 centiMorgans (cM) include recombinant and parental gametes. With a distance of 15 cM, we can expect approximately 15% of the gametes to be recombinants and 85% to remain parental types.
The parental gametes will have the combinations aB and Ab from the linked alleles on the chromosome, and the recombinant gametes will have the combinations AB and ab, which are new allele combinations formed through crossovers during meiosis.