This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: = 0.2 a = q 2 + 1/2 (2pq) We could also do this by calculating a first: a = 1A = 10.2 = 0.8 Since the freq of A and a must add up to 1, we can calculate a by subtraction: So in this case, a = 0.64 + 1/2 (0.32) = 0.8 2) Given HW equilibrium and the frequency of an allele , what % of the population will have various genotypes ? Genotypes are given by p 2 + 2pq+ q 2 , so: From #1: A = 0.2 a = 0.8 i) What is the frequency of homozygous AA individuals? AA => p 2 = 0.2*0.2 = 0.04 ii) What is the frequency of homozygous aa individuals? aa => q 2 = 0.8*0.8 = 0.64 i) What is the frequency of heterozygous Aa individuals? Aa => 2pq = 2*0.8*0.2 = 0.32...
View
Full
Document
This note was uploaded on 11/22/2009 for the course BIO SCI 150 taught by Professor Geraldbergstrom during the Fall '08 term at Wisconsin Milwaukee.
 Fall '08
 GeraldBergstrom
 Genetics, Population Genetics

Click to edit the document details