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Super Smash Brothers 4 for Uchenna !

Super Smash Brothers 4 for Uchenna !. Warm Up! 1/21/14. What evidence do we have that evolution is occurring? Name 3 things. Take 1 answer from question #1 & describe how that provides evidence for evolution. Remember! Populations & Gene Pools.

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Super Smash Brothers 4 for Uchenna !

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  1. Super Smash Brothers 4 for Uchenna!

  2. Warm Up! 1/21/14 • What evidence do we have that evolution is occurring? Name 3 things. • Take 1 answer from question #1 & describe how that provides evidence for evolution.

  3. Remember! Populations & Gene Pools • A gene pool is a collection of alleles in the population. • Don’t forget the difference between allele & gene! • The allele frequency is how common that allele is in the population. Frequency of A? 12/20 or 60% Frequency of a? 8/20 or 40% A = gray, a = white This gene pool consists of 20 alleles.

  4. How do we know if a population is evolving? • We study evolution as genetic change by comparing it to a population that is NOT evolving. • There are 5 criteria that a population needs to meet in order to be NON-evolving. • This is the Hardy-Weinberg Principle. • Populations that meet these 5 criteria are said to be in Hardy-Weinberg Equilibrium.

  5. Hardy-Weinberg Equilibrium

  6. The 5 Criteria of Hardy-Weinberg EquilibriumTo be considered a non-evolving population… • Random Mating: all members of the population have an equal shot at mating; selecting mates based on traits does not occur

  7. The 5 Criteria of Hardy-Weinberg EquilibriumTo be considered a non-evolving population… 2. No Genetic Drift: GD is a random change of frequency of an allele due to a small population size • Natural selection affects small populations much more than large ones! 3. No immigration or emigration: no new individuals can add to the gene pool

  8. The 5 Criteria of Hardy-Weinberg EquilibriumTo be considered a non-evolving population… 4. No mutations: these are the ultimate source of variation, so a non-evolving population wouldn’t have mutations 5. No natural selection: all individuals have the same ability to survive & reproduce

  9. What are some examples of populations that are in Hardy-Weinberg Equilibrium? DOES NOT EXIST IN NATURE!

  10. Hardy-Weinberg Equilibrium • Hypothetical, non-evolving population • Allele frequencies stay constant • Serves as a model (a null hypothesis) • We can measure evolutionary change in a population by comparing it a population that’s in HW Equilibrium (not evolving). W. Weinberg, physician G.H. Hardy, mathematician

  11. The Hardy-Weinberg Theorem – Counting Alleles Assume there are 2 alleles: B & b Frequency of the dominant allele (B): p Frequency of the recessive allele (b): q Frequencies must add to 1 (100%), so: p + q = 1

  12. The HW Theorem – Counting Alleles p = .6 q = .4 p + q = 1 Frequency of A: 12/20 or 60% Frequency of a: 8/20 or 40% A = gray, a = white This gene pool consists of 20 alleles.

  13. The HW Theorem – Counting Individuals Frequency of Homozygous Dominant: p x p = p2 Frequency of Homozygous Recessive: q x q = q2 Frequency of Heterozygous: (p x q) + (q x p) = 2pq Frequencies must add to 1 (100%), so: p2 + 2pq + q2 = 1

  14. The HW Theorem – Counting Individuals Frequency of homozygous dominant: p2 = .36 Frequency of homozygous recessive: q2 = .16 Frequency of heterozygous: 2pq = .48 Total: p2 + 2pg + q2 = 1 p = .6 q = .4 p + q = 1

  15. HW Formulas • Alleles: p + q = 1 • Individuals: p2 + 2pg + q2 = 1

  16. In this population of pigs, the allele for pink coat (B) is dominant & the allele for black coat (b) is recessive. What is p? What is q? 16 pigs total! 32 alleles total! If q = .5, then p = .5 (because p + q = 1) Total black (bb) pigs = 4 SO, q2 = 4/16 = .25  q = .5

  17. In this population of pigs, the allele for pink coat (B) is dominant & the allele for black coat (b) is recessive. What percent of the pig population is heterozygous? The frequency of heterozygotes = 2pg = .5 = 50% Remember, p = .5 & q = .5

  18. Using the HW Equation • Out of 100 cats, 84 are black & 16 are white. How many of each genotype? q2 (bb) = 16/100 = .16 q (b) = .4 Now we can find the frequency of cats of each genotype! Since we know p + q = 1… 1 – q = p (B) = .6 p2 (BB) = .36 2pg (Bb) = .48 q2 (bb) = .16

  19. Using the HW Equation To determine if this population is in HW equilibrium, the next generation will have the same frequencies, & the next generation, & the next generation… p2 (BB) = .36 2pg (Bb) = .48 q2 (bb) = .16 Total = 1

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