Final answer:
The expected frequency of Ab gametes from an F1 heterozygous individual AaBb when loci 'a' and 'b' are 10 cM apart is 5%, since this represents the recombinant gametes due to a 10% recombination frequency between the loci.
Step-by-step explanation:
The expected frequency of Ab gametes produced from the heterozygous F1 individual with genotype AaBb, when loci "a" & "b" are 10 cM apart, can be found using the concept of genetic linkage and recombination. In this case, 10 cM (centiMorgans) distance indicates that there is a 10% chance of recombination between the two loci per generation. Thus, since the loci are partially linked, the expected frequency of the recombinant Ab gamete would be 10% of the total gametes produced.
Under Mendelian inheritance with no linkage, you would expect a 1:1:1:1 ratio of the gametes (AB, Ab, aB, ab). However, the presence of linkage alters the expected ratios. Parental gametes (AB and ab) are more frequent than recombinant gametes (Ab and aB) due to the linkage between the loci. The exact proportion of the recombinant gametes depends on the frequency of crossing over events that occur between the linked genes during meiosis.
Given that Ab is a recombinant genotype, resulting from recombination, we can expect 5% of the gametes to be Ab and another 5% to be aB, reflecting the 10% recombination frequency. Therefore, out of 100 gametes, we can expect approximately 5 to be of the Ab type.