**Problem 3: Genetic Linkage and Mapping**In Drosophila flies, the allele *b* results in a black body, while the allele *b+* leads to a brown body, the wild-type phenotype. The allele *wx* indicates waxy wings, while *wx+* represents non-waxy wings, the wild-type phenotype. Meanwhile, the allele *cn* causes cinnabar eyes, and *cn+* leads to red eyes, the wild-type phenotype.In this experiment, a female fly heterozygous for these three genes is testcrossed. The progeny are categorized as follows out of 1000 offspring:- 5 wild type- 6 black, waxy, cinnabar- 69 waxy, cinnabar- 67 black- 382 cinnabar- 379 black, waxy- 48 waxy- 44 black, cinnabarThe task is to determine if these three genes are linked. If they are, you are to draw a chromosome map for the heterozygous parent, indicating the order, arrangement, and map distance between the linked genes.

If the genes are located on the same chromosome then they are classified as linked genes and the crossing over frequency between the genes will be reduced. This crossover frequency is used to calculate the distance between the genes.Answer:To determine if the genes are linked, we first examine the progeny phenotypes. The two most common phenotypes are "cinnabar" and "black, waxy", which likely represent the parental combinations. This suggests that the genes are linked, with the arrangement being either b-wx-cn or cn-wx-b.To determine the order of the genes, we look at the double crossover progeny, which are the least common phenotypes. These are "waxy" and "black, cinnabar". Comparing these phenotypes with the parental phenotypes, we can see that the waxy phenotype has changed, suggesting that the wx gene is in the middle. So the arrangement is b-wx-cn.The recombination frequencies to determine the distances between the genes:The map of the chromosomes would be:b------53.8------wx------54.3------cnNote: The map distances are approximations and might not add up perfectly due to factors like crossover interference.