Hardy-Weinberg - Alkaptonuria 1 in 100,000 1 in 503...

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Gene 3000 03-06-07 Insert between slides 12 and 13 Application of the Hardy-Weinberg equation For genetic disorders controlled by a recessive allele: o Calculate the frequency of carriers (heterozygotes) from the frequency of affected individuals (homozygotes) Disorder Affected Carriers Albinism 1 in 20,000 1 in 72 Phenylketonuria 1 in 75,000 1 in 80
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Unformatted text preview: Alkaptonuria 1 in 100,000 1 in 503 Albinism q 2 = 1/20,000 = 0.00005 q = 0.0071 p = 1-q = 0.9929 2pq = carriers = 0.0141 More on Hardy-Weinberg How do we test for fitness between o Observed genotypic frequencies o Expected frequencies under Hardy-Weinberg Use statistical tests such as chi 2 Data from nature 1...
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This note was uploaded on 04/17/2008 for the course GENE 3000 taught by Professor Arnold/anderson during the Spring '07 term at University of Georgia Athens.

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