Final answer:
The expected frequency of the Ab gamete from the F1 heterozygous individual (AaBb) with loci 'A' and 'B' that are 2.8 cM apart is 2.8%, corresponding with the recombination frequency for the genetic distance between these loci.
Step-by-step explanation:
The expected frequency of the Ab gamete produced from a heterozygous F1 individual (AaBb) when loci 'A' and 'B' are 2.8 cM apart is calculated by considering the recombination frequency corresponding to the genetic distance between the loci. Since the loci are 2.8 centimorgans (cM) apart, the recombination frequency is 2.8%, representing the likelihood of crossover during gamete formation.
Given that the homozygous dominant parent is AABB, and the homozygous recessive parent is aabb, the F1 generation will be AaBb. During meiosis, the alleles A and a will segregate, as will B and b. Because the genes are linked, there will be a higher frequency of parental-type gametes (AB and ab) compared to recombinant gametes (Ab and aB). However, due to crossovers, recombinant gametes can still occur. According to the linkage and the recombination frequency, the proportion of Ab gametes is expected to be 2.8%. This frequency of recombination is what we would expect on average if the loci are as per the provided genetic distance apart.