The genes for sepia eye color, short bristles, and dark body coloration are on the same chromosome of *Drosophila melanogaster*. Each gene has two alleles: wild type, which is dominant, or mutant, which is recessive.- **se+** is dominant and causes red eyes; **se** is recessive and causes sepia eyes.- **sb+** is dominant and causes long bristles; **sb** is recessive and causes short bristles.- **b+** is dominant and causes gray body coloration; **b** is recessive and causes dark body coloration.The **sepia** gene and **short bristle** gene are separated by 33 map units; the **short bristle** gene and the **dark body** gene are separated by 12 map units. The **sepia** gene and **dark body** gene are separated by 45 map units.If you crossed a fly, which was homozygous for the wild type allele of each gene (and had the phenotypes red eyes, long bristles, and gray body) with a homozygous recessive fly for each gene (and had the phenotypes sepia eyes, short bristles, and dark body coloration), the F1 generation would entirely consist of flies that had the wild type traits, but were heterozygous for each gene.If you were to conduct a test cross, and cross these heterozygous F1 flies to recessive homozygous flies, how many flies would you predict to observe in each of the eight predicted phenotypic categories with respect to these three genes (given the map units above) if there were **1,000 total offspring**? Show your work. (Some portions are marked with black lines.)

Name: The genes for sepia eye color, short bristles, and dark body colorati
Uploaded: 2024-03-20T06:43:04.000Z
Channel: Bartleby
Description: The genes for sepia eye color, short bristles, and dark body coloration are on the same chromosome of *Drosophila melanogaster*. Each gene has two alleles: wild type, which is dominant, or mutant, which is recessive.- **se+** is dominant and causes

To determine the expected phenotypic ratios in the F₂ generation of a test cross, you can use the information about gene linkage and the distances between the genes.Let's denote the wild-type alleles as "+", and the mutant alleles as the lowercase form of the gene symbol.The parental flies are:- P₁ (parent 1): se+ sb+ b+ (red eyes, long bristles, gray body)- P₂ (parent 2): se sb b (sepia eyes, short bristles, dark body)Given that the sepia gene and short bristle gene are separated by 33 map units, the short bristle gene and dark body gene are separated by 12 map units, and the sepia gene and dark body gene are separated by 45 map units.P₁: se+ sb+ b+ |______| (33 map units) |______| (12 map units) |______|P₂: se sb bNow, let's consider the recombination events that can occur during meiosis in the F₁ heterozygous flies (se+ sb+ b+/se sb b). The recombination frequencies are given by the map units.Recombination between sepia and short bristle genes (33 map units):Parental type: se+ sb+ b+ / se sb bRecombinant type: se+ sb b / se sb+ b+Recombination between short bristle and dark body genes (12 map units):Parental type: se+ sb+ b+ / se sb bRecombinant type: se+ sb b / se sb+ b+Recombination between sepia and dark body genes (45 map units):Parental type: se+ sb+ b+ / se sb bRecombinant type: se+ sb b / se sb+ b+Therefore, we would predict to observe the following number of flies in each of the eight predicted phenotypic categories:Red eyes, long bristles, gray body: 375 fliesSepia eyes, long bristles, gray body: 187.5 fliesRed eyes, short bristles, gray body: 62.5 fliesSepia eyes, short bristles, gray body: 31.25 fliesRed eyes, long bristles, dark body: 31.25 fliesSepia eyes, longbristles, dark body: 1, short bristles, dark body: 7.8125 flies