Danielle 7

Danielle 7 - Homework Required Problems: 7.1, 7.2, 7.3,...

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Homework Required Problems: 7.1, 7.2, 7.3, 7.7, 7.8, 7.10 (test w/ χ 2), 7.11 (test w/ χ 2), 7.12 (test w/ χ 2), 7.13, 7.14, 7.16, 7.17, 7.18, 7.19, 7.21, 7.23, 7.24, 7.29. Plus additional questions to be posted. Due March 5 or 7.
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Ch. 7 Linkage, recombination, & chromosome mapping Recall that crossing-over occurs in prophase of Meiosis I. Crossing-over involves exchange of DNA between chromosomes
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Meiosis (No crossing-over) Consider genes A and B on same chromosome, without crossing-over: A B A B A B a b a b a b Parental allele arrangement: A B/a b Only produce AB or ab gametes (parental gametes) This isn’t sufficient variation for natural selection, evolution, etc.
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Genes on same chromosome often do not assort independently. Need new notation to show which alleles on same chromosome. A B/a b Alleles on different chromosome homologs Crossing over does allow some assortment of genes.
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Meiosis (with crossing-over) With crossing-over: A B A B A b a b a B a b Non-recomb. (Parental) Non-recomb. (Parental) Recombinant (from crossing- over) Four types of gametes: AB, Ab, aB, and ab Notice how here each gamete is different. desired for evolution and natural selection.
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Genetic map distance (recombination frequency) = # of recombinant gametes Total gametes X 100 Genetic map distance measured in centiMorgans (cM). A recombination frequency of 1% has historically been called one map unit since the time of the Sturtevant and Morgan. Although the term "map unit" is still widely used, one map unit is now often referred to as one centimorgan , honoring Thomas Hunt Morgan. Calculating recombination frequency
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Crossing over If the genes are not on the same chromosome, independent assortment will occur and the test cross will yield a 1:1:1:1 ratio of all four possible phenotypes. In other words, in a dihybrid cross, departure from a 1:1:1:1 ratio of F1 gametes indicates that the two genes are not on the same chromosome. If it happens this way, about ½ of the gametes will be of the two parental types, and the other ½ will be of the 2 recombinant types. For linked genes, the probability of crossing over increases with physical distance between the two genes. If recombination occurs, some of the progeny of the test cross will no longer exhibit the parental phenotypes. You can see if the 1:1:1:1 ratio of the 4 different kinds of gametes actually materializes by counting the different types of male progeny in the F2 generation. The percentage of recombinants detected in the progeny of the test cross increases with the distance between the genes, and is referred to as map distance .
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For example…. Consider a test cross of a female of genotype Ab/aB that yields the following phenotypes: 495 Ab 495 aB 5 AB 5 ab. In a total of 1000 progeny, there are 5 AB and
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Danielle 7 - Homework Required Problems: 7.1, 7.2, 7.3,...

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