Please explain your answer and show calculations thanks For

Please explain your answer and show calculations, thanks!

For questions 1-6. fluff is a mutation in Drosophila melanogaster with short fine bristles. Carmine eyes is a mutation that has a different color than the normal brick-red. And ocellarless is a mutation with the ocellars missing. A Drosophila male with ocellarles is crossed to a female with carmine eyes and fluff bristles. The F1 males all have carmine eyes. The F1 females are all ocellarless F1 females with ocellarles are crossed males that are homozygous or hemizygous for the recessive allele at each locus segregating in the F1 female and produce the following progeny: Which mutant allele(s), if any is are dominant to the respective normal allele?______ Which mutant gene allele(s). If any is are X-lined?______ How many loci are segregating (that is, heterozygous) in the F1 female?_______ What is the Chi-square value and the degree of freedom for the data shown above?_______ If any genes are linked, what is the gene order?_______ What is the total length of the map in question = 5?________

Solution

1. Carmine eyes and Ocellarless are the mutant alleles that are dominant to the respective normal allele.

Because when a cross is made between Ocellarless male and carmine eyes female, all the female f1 progeny showed Ocellarless and all the f1 male progeny showed carmine eyes. Therefore by this we can say that these two are the dominant alleles.

3. Three loci ( fluff, carmine, Ocellarless)