A) The male parent is black with normal wings, so its genotype is "BBNN" (uppercase letters indicate dominant alleles).
B) they all have the genotype "BbNn".
C) the male's genotype is "bbnn".
D) The expected distribution of genotypes would be: 400 brown wingless flies (genotype "bbnn"), 400 black winged flies (genotype "BBNN"), 400 brown wingless flies (genotype "bbNn"), and 400 black wingless flies (genotype "BBnn").
E) the genetic distance between the color and wing genes is (813 / 1600) x 100 = 50.81%.
A) The P generation consists of two true-breeding flies. The female parent is brown and wingless, which means it has the genotype "bbnn" (lowercase letters indicate recessive alleles). The male parent is black with normal wings, so its genotype is "BBNN" (uppercase letters indicate dominant alleles).
B) The F1 generation results from crossing the P generation flies. All the flies in the F1 generation are brown and have normal wings. This means that they all have the genotype "BbNn" (one dominant and one recessive allele for each trait).
C) To find the genotype of the male parent in the next cross, we know that it is black with wingless wings, so it must have the recessive alleles for both traits. Therefore, the male's genotype is "bbnn".
D) If the wing and color traits were linked and no recombination occurred, we would expect to count 1,600 offspring in the F2 generation. The expected distribution of genotypes would be: 400 brown wingless flies (genotype "bbnn"), 400 black winged flies (genotype "BBNN"), 400 brown wingless flies (genotype "bbNn"), and 400 black wingless flies (genotype "BBnn").
E) However, when you count the actual F2 generation, you get different numbers. There are 85 brown wingless flies, 728 black winged flies, 712 brown wingless flies, and 75 black wingless flies. To calculate the genetic distance between the color and wing genes, we can use the formula: genetic distance = (number of recombinant offspring / total number of offspring) x 100.
In this case, the number of recombinant offspring is the sum of the brown wingless flies (85) and the black winged flies (728), which is 813. The total number of offspring is 85 + 728 + 712 + 75, which is 1600.
So, the genetic distance between the color and wing genes is (813 / 1600) x 100 = 50.81%.