Final answer:
The correct answer to the two-point testcross in Drosophila with a map distance of 10 cM between genes is b. 1:1:1:1. This represents the phenotypic ratio of offspring considering the likelihood of recombination between linked genes.
Step-by-step explanation:
When performing a two-point testcross in Drosophila with a map distance of 10 cM between two genes, the expected phenotypic ratio following independent assortment would normally be 9:3:3:1, as seen in a Mendelian dihybrid cross with two non-interacting genes. However, the presence of a 10 cM map distance implies that these genes are linked and that there will be recombination between them. As a result, we would expect approximately 10% of the offspring to exhibit recombination phenotypes.
The ratio provided in the options suggests correction for linkage, which will affect the outcome of the expected phenotypic ratio. Given the calculated map distance of 10 cM, we can infer that 10% of the offspring will be recombinant types, leading to an expected ratio of 1:1:1:1 (parental types and recombinant types will both occur at roughly equal frequency). Therefore, the correct answer is b. 1:1:1:1, which reflects the expected ratio when considering recombination events between linked genes.