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Propose a possible reason that the first naturally occurring mutant fruit fly Morgan saw involved a gene on a sex chromosome and was found a male.

Short Answer

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When the gene is located on the sex chromosome, male Drosophila needs one mutant allele to exhibit a mutant phenotype. However, if the genes are present on autosomes, two mutant alleles are required to produce a mutant phenotype.

Step by step solution

01

Autosomes and sex chromosomes

A chromosome is the discrete unit of the genome that carries genes.These are self-replicating molecules involved in heredity, mutation, evolution, variation, etc. Autosomes and sex chromosomes are the two types of chromosomes.

Genes that determine sex-linked traits are present on sex chromosomes, while genes that determine all other traits of an individual are present on autosomes.

02

Fruit fly

Drosophila melanogasteris commonly known as the fruit fly. It is a small insect and feeds on fungi. This insect has four pairs of chromosomes; one pair of sex chromosomes and three pairs of autosomes.

Male Drosophila possesses one X and one Y chromosome, while females possess two X chromosomes. This fly produces many offspring in a single mating and has a short generation time. Thus, these insects are easy to grow in the lab.

03

Reason for the occurrence of the first mutation in sex chromosome

The first mutant Drosophila was a male with white eyes.However, the wild type has red eyes.The gene for the eye color of Drosophila is located on X-chromosome.

Thus, a male with a defective X chromosome could exhibit the mutant phenotype as males possess one X and one Y chromosome.

However, if the gene linked to the mutant phenotype was present on autosomes, then both the alleles of the gene had to be mutant to exhibit the phenotype.This is quite a rare situation.

Thus, mutant phenotype linked to sex chromosome in male fruit fly was important to identify the mutant organism.

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Most popular questions from this chapter

The gene that is activated on the Philadelphia chromosome codes for an intracellular tyrosine kinase. Review the discussion of cell cycle control in Concept 2.3 and explain how the activation of this gene could contribute to the development of cancer.

Assume you are mapping genes A, B, C, and D in Drosophila. You know that these genes are linked on the same chromosome, and you determine the recombination frequencies between each pair of genes to be as follows: A-B, 8%; A-C, 28%; A-D, 25%; B-C, 20%; B-D, 33%.

  1. Describe how you determined the recombination frequency for each pair of genes.

  2. Draw a chromosome map based on your data.

A wild-type fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings. The offspring have the following phenotypic distribution: wild type, 778; black vestigial; 785; black normal, 158; gray vestigial, 162. What is the recombination frequency between these genes for the body color and wing size? Is this consistent with the results of the experiment in Figure 15.9?

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Using the information from problem 4, scientists do a further testcross using a heterozygote for height and nose morphology. The offspring are tall upturned snout, 40; dwarf upturned snout, 9; dwarf downturned snout, 42; tall downturned snout, 9. Calculate the recombination frequency from these data, and then use your answer from problem 4 to determine the correct order of the three linked genes.

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