Chapter 15 notes - Chapter 15 Chromosomal Basis of...

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Chapter 15 Chromosomal Basis of Inheritance Thomas Hunt Morgan first to associate a specific gene with a specific chromosome in the early 20th century. wild type : The normal character phenotype mutant phenotypes: Alternative traits Deduced that the gene with the white-eyed mutation is on the X chromosome alone, a sex-linked gene. Females (XX) may have two red-eyed alleles and have red eyes or may be heterozygous and have red eyes. Males (XY) have only a single allele and will be red eyed if they have a red- eyed allele or white-eyed if they have a white-eyed allele. Linked genes: Genes located on the same chromosome that tend to be inherited together because the chromosome is passed along as a unit. Results of crosses with linked genes deviate from those expected according to independent assortment. Morgan’s Experiment He crossed F1 heterozygous females (b+bvg+vg) with homozygous recessive males (bbvgvg). According to independent assortment, this should produce 4 phenotypes in a 1:1:1:1 ratio. He observed a large number of wild-type (gray-normal) and double-mutant (black-vestigial) flies among the offspring. These phenotypes correspond to those of the parents. Morgan reasoned that body color and wing shape are usually inherited together because their genes are on the same chromosome. The other two phenotypes (gray-vestigial and black-normal) were fewer than expected from independent assortment and totally unexpected from dependent assortment. These new phenotypic variations must be the result of crossing over. Genetic recombination: The production of offspring with new combinations of traits inherited from two parents
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can result from independent assortment of genes located on nonhomologous chromosomes or from crossing over of genes located on homologous chromosomes. Parental types : organisms with phenotypes that match the original P parents Recombinants : new combination of parental traits Morgan’s Results The results of Morgan’s testcross for body color and wing shape did not conform to either independent assortment or complete linkage. Under independent assortment the testcross should produce a 1:1:1:1 phenotypic ratio. If completely linked, we should expect to see a 1:1:0:0 ratio with only parental phenotypes among offspring.
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