Genes for body color (B black dominant to b yellow) and wing shape (C straight dominant to c curved) are located on the same chromosome in flies. If single mutants for each of these traits are crossed (i. e. a yellow fly crossed to a curved-wing fly), and their progeny is testcrossed, the following phenotypic ratios are observed among their progeny. Black, straight 17
yellow, curved 12
black, curved 337
yellow, straight 364

Required:
Calculate the map distance between B and C.

Map distance = Genetic distance, GD = 3.9 MU ≅ 4 MU

Explanation:

The more separated two genes are, the more chances of recombination there will be.  The closer they are, the fewer chances of recombination there will be. Two genes that are very close will have a few recombination events and are strongly bounded.    

To analyze the recombination frequency, we have to know that:

The map unit is the distance between the pair of genes for which every 100 meiotic products one of them results in a recombinant one.  

Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to identify the genotypes of the parental gametes with the ones of the recombinants. We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the recombinants are the less frequent. So:

Parentals:

Recombinants:

To calculate the recombination frequency we will make use of the next formula: P = Recombinant number / Total of individuals.

P = Recombinant number / Total of individuals.  

P = 17 + 12 / 337 + 364 + 17 + 12

P = 29 / 730

P = 0.039

The genetic distance (GD) will result from multiplying that recombination frequency (P) by 100 and expressing it in map units (MU).

GD = P x 100

GD = 0.039 x 100

GD = 3.9 MU ≅ 4 MU

i believe it would cause an increase of percipitation, though i could be wrong. so i would say b is the answer.