This preview shows page 1. Sign up to view the full content.
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...
View Full Document
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.
- Spring '07