Hardy Weinberg Populations

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Transcript of Hardy Weinberg Populations

  • 1.Measuring Evolution of Populations

2. 5 Agents of evolutionary change Mutation Gene Flow Genetic Drift Selection Non-random mating 3. Populations & gene pools

  • Concepts
    • apopulationis a localized group of interbreeding individuals
    • gene pooliscollection of allelesin the population
      • remember difference betweenalleles(gene variations) &genes !
    • allele frequency is how common that allele is within the population
      • how manyAsvs.asin whole population

4. Evolution of populations

  • Evolution =change in allele frequenciesin a population
    • hypothetical : what conditions would cause allele frequencies tonotchange?
    • non-evolving population
      • REMOVEall agents of evolutionary change
      • very large population size (nogenetic drift )
      • no migration (nogene flowin or out)
      • nomutation(no genetic change)
      • random mating(no sexual selection)
      • nonatural selection(everyone is equally fit)

5. Hardy-Weinberg equilibrium

  • Hypothetical, non-evolving population
    • preserves allele frequencies
  • Serves as a model
    • natural populations rarely inH-W equilibrium
    • useful model tomeasureif forces are
    • acting on a population
      • measuring evolutionary change

W. Weinberg physician G.H. Hardy mathematician 6. Hardy-Weinberg theorem

  • CountingAlleles
    • assume 2 alleles =B ,b
    • frequencyof dominant allele ( B ) = p
    • frequencyof recessive allele ( b ) =q
      • frequencies must add to 1 (100%), so:
    • p+q= 1

bb Bb BB 7. Hardy-Weinberg theorem

  • CountingIndividuals
    • frequency ofhomozygous dominant :pxp=p 2
    • frequency ofhomozygous recessive : qxq=q 2
    • frequency ofheterozygotes :( pxq ) + ( qxp ) =2p q
      • frequencies ofall individualsmust add to 1 (100%), so:
    • p 2+ 2pq +q 2= 1

bb Bb BB 8. H-W formulas

  • Alleles: p +q = 1
  • Individuals: p 2+ 2p q + q 2= 1

bb Bb BB BB B b Bb bB bb 9. Using Hardy-Weinberg equation What are the genotype frequencies? q 2(bb): 16/100 = .16 q(b): .16 =0.4 p(B): 1 - 0.4 =0.6 population:100 cats 84 black, 16 white How many of each genotype? bb Bb BB p 2 =.36 2pq =.48 q 2 =.16 Must assume population is in H-W equilibrium! 10. Any Questions?? Any Questions?? Any Questions??