Question

You are doing a genetics experiment with the fruit fly. In the "P" generation, you cross two true-breeding flies. The female parent is brown and wingless and the male parent is black with normal wings. All of the flies in the F1 generation are brown and have normal wings. Indicate the alleles associated with dominant phenotypes by uppercase letters and alleles associated with recessive phenotypes by lowercase letters. Assume the genes are not found on a sex chromosome. Indicate the color alleles as "B" and "b" the wing alleles by the letters "N" and "n a) What are the genotypes in P generation? b) What are the genotypes in F1 generation? c) You now take an F1 female and cross her to a true-breeding black, wingless male. and you count 1600 offspring in the F2 generation and you get: 85 brown winged flies 728 black winged flies 712 brown wingless flies 75 black wingless flies These results suggest the genes are on the same chromosome, why? d) What is the genetic distance between the color and wing genes? e) A series of fruit fly matings shows that the recombination frequency between the gene for wing size and the a third gene (the gene for antenna length is 5% . List all possible recombination frequencies between the gene for color and the gene for antenna length and draw the possible chromosome map(s).
Drosophila genetics hw. Can someone help explain this

Answer 1

According to the information provided, key representing the allele information are:

B : Brown body (dominant phenotype)

b : Black body (recessive phenotype)

N : Normal wings (dominant phenotype)

n : Wingless (recessive phenotype)

(a) Since the parental generation comprises of true-breeding flies, the parental generation is homozygous for both the genes.Thus, the genotype of the parents are

Female fly : BBnn (brown body and wingless)

Male fly : bbNN (black body and normal wings)

(b) Upon crossing, the genotype of the F1 generation progeny : BbNn (Brown body & normal wings)

(c) According to the question, the crossing occurs betwen BbNb (female) X bbnn (male) and the number of progenies based on the phenotype are presented in the table.

Genetic markers located on the same chromosome tend to remain together during sexual reproduction and a test cross ratio which ideally produces 50% progeny with parental and 50% progeny with recombinant phenotypes will be altered. As in the given condition, only 10% parental phenotypes are obtained clearly highlight that the genes are located on the same chromosome and do not follow the law of independent assortment.

(d) From the frequency of the homozygous recessive genotype bbnn, this genotype can be formed only when a bn female gamete is fertilized with bn male gamete. The frequency of teh bn gamete type is, thus, equal to the square root of the frequency of the double homozygous recessive genotype. From the data,

the frequency of bbnn genotype = 75/1600 = 0.04

Thus, the frequency of bn gamete equals $\sqrt{}$ 0.04 = 0.2

Thus, the recombination frequency equals 2(0.2) i.e. 0.4 and hence the b and n loci are 40 map units apart.