Question

8. You are doing a genetics experiment with the fruit fly. In the P generation, you cross 2 true-breeding flies. The female parent is brown and wingless and the male parent is black with normal wings. All the flies in the F1 are brown and have normal wings. Indicate the alleles associated with dominant phenotypes by uppercase letters and the alleles associated with recessive phenotypes by lowercase letters. Assume that the genes are not found on sex chromosomes. Indicate the color alleles as " B " and " b " the wing alleles by the letters " N " and " n" (A) What are the genotypes of the P generation of each parent? (B) What is the genotype of the F1 generation? (C) You now take an F1 female and cross her to a true-breeding black, wingless male. What is this male's genotype? (D) You count 1,600 offspring in the F2 generation. If the wing and the color traits were linked and no recombination occurred, you would expect to count: of brown wingless flies (Genotype is of black, winged flies (Genotype is of brown, wingless flies (Genotype is of black, wingless flies (Genotype is (E) When you count the F2 generation, you really get: 85 brown wingless flies 728 black winged flies 712 brown wingless flies 75 black wingless flies What is the genetic distance between the color and wing genes?

Answer 1

In a genetics experiment with fruit flies, the F1 generation brown with normal wings indicates dominance of these traits. The P generation genotypes are bbnn and BBNN, while the F1 genotype is BbNn. Recombinants in the F2 generation suggest a genetic distance between the color and wing genes.

Step-by-step explanation:

You are doing a genetics experiment with fruit flies (Drosophila melanogaster). When crossing two true-breeding flies, one brown and wingless, and the other black with normal wings, all F1 flies are brown with normal wings. This suggests brown color and normal wings are dominant traits, while black color and winglessness are recessive traits.

(A) The genotypes of the P generation for each parent are bbnn for the female (brown and wingless) and BBNN for the male (black with normal wings), if we assume brown and normal wings are dominant.

(B) The genotype of the F1 generation is BbNn since all flies are brown with normal wings.

(C) A true-breeding black, wingless male would have the genotype bbnn.

(D) If the wing and the color traits were linked and no recombination occurred, we would expect a 1:1:1:1 ratio of the genotypes in the F2 generation.

(E) The actual count of the F2 generation shows a significant deviation from the expected 1:1:1:1 ratio, indicating recombination. The genetic distance between the color and wing genes can be calculated using the recombination frequency: (Number of recombinants / Total offspring) × 100. In this case, recombinants are brown wingless and black wingless flies. So, the calculation is ((85+75)/1600)× 100, which gives the genetic distance between the color and wing genes.

Answer 2

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%.