Final answer:
The precise map distance calculated from the testcross data is neither of the provided options, as the actual map distance is 4.17 cM, which equates to the recombination frequency observed among the progeny after excluding the lethal genotype.
Step-by-step explanation:
In order to calculate the most precise map distance from the data given in the testcross of AB/ab flies crossed to ab/ab flies, we consider only the progeny that resulted from recombination. Since the genotype Aa/bb is lethal and therefore does not appear among the progeny, we have to exclude these from our calculations. The recombination progeny are the ones with non-parental types, which are aa Bb and Aa bb. However, since Aa bb is lethal and doesn't appear, we only count aa Bb which are 20 in number.
The total number of progeny is the sum of all progeny, which is 235 + 225 + 20 = 480. To find the recombination frequency, which corresponds to the map distance, we take the number of recombinants (20) and divide by the total number of progeny (480), then multiply by 100 to get a percentage:
Recombination frequency = (20/480) * 100 = 4.17%.
According to Sturtevant's method of using recombination frequency to calculate map distance in centimorgans (cM), we can equate 1% recombination frequency to 1 cM. Therefore, the map distance would be 4.17 cM. Looking at the answer options provided, the closest given value to 4.17 cM is 8.0 cM. However, it seems there might be a mistake, as the actual calculated value is not among the options. The exact calculated map distance is neither 8.0 cM, 10.0 cM, 12.5 cM, nor 15.0 cM, based on the calculated recombination frequency of 4.17%.