text: Hardy-Weingberg practice questions
  
  # Hardy-Weingberg practice questions - = 0.2 a = q 2 + 1/2...
  
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  Review: Population Genetics Problems Two types of questions: 1) Given H-W equilibrium and the frequency of various genotypes (e.g., AA, Aa or aa), what % of the population will have the allele A (or a)? 2) Given H-W equilibrium and the frequency of an allele (eg. A or a) , what % of the population will have genotypes AA, Aa or aa?
  
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  1) Given H-W equilibrium and the frequency of genotypes , what is the frequency of each allele ? For example, let's say we have the following genotype frequencies: AA =0.04 Aa = 0.32 aa = 0.64 A = p 2 + 1/2 (2pq) What is the frequency of the A allele ? p 2 q 2 2pq Under H-W, the frequency of these genotypes is given by: The general formula for determining the frequency of A is: So in this case, A = 0.04 + 1/2 (0.32)
  
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  Unformatted text preview: = 0.2 a = q 2 + 1/2 (2pq) We could also do this by calculating a first: a = 1-A = 1-0.2 = 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 H-W 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...
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  ## 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.
  
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  Hardy-Weingberg practice questions - = 0.2 a = q 2 + 1/2...
  
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