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Measuring Evolution of Populations

Learn about the conditions for Hardy-Weinberg equilibrium, the Hardy-Weinberg equation, and how to calculate allele and genotype frequencies. Explore the concept of evolution in populations.

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Measuring Evolution of Populations

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  1. MeasuringEvolution of Populations Image from: https://s-media-cache-ak0.pinimg.com/236x/0a/c4/6c/0ac46cac37e5afb70010f25beedec39e.jpg

  2. CONDITIONS FOR HARDY WEINBERG EQUILIBRIUM http://pngimg.com/upload/hands_PNG905.png

  3. Populations & gene pools _________ = group of interbreeding individuals ___________is collection of alleles in the population • remember difference between alleles & genes! allele ___________is how common is that allele in the population • how many A vs. a in whole population

  4. Hardy-Weinberg equilibrium POPULATION IN HW EQUILIBRIUM = IDEAL IT MEANS NO EVOLUTION IS HAPPENINGAllele frequencies stay the sameRare in real populations Way to tell if evolution is happening in a population G.H. Hardy mathematician W. Weinberg physician

  5. The Hardy-Weinberg Equation p2 + 2pq + q2 = 1 _____ = the frequency of homozygous dominant genotype (T T) ______ = the frequency of heterozygous genotype (T t) ______ = the frequency of homozygous recessive genotype (t t)

  6. The Hardy-Weinberg Equation p + q = 1 _____ = the frequency of dominant ALLELE in population (T) ______ = the frequency of recessive ALLELE in population (t) BOZEMAN BIOLOGY Hardy Weinberg equation

  7. T t T p + q = 1 t ALLELES in population T = _____ t = _____

  8. T t T T t TT Tt tt t Homozygous dominant = ______ Homozygous recessive = ______ Heterozygous = _____

  9. T t T p2 + 2pq + q2 = 1 t GENOTYPES in population Homozygous dominant = ________ Homozygous recessive = ________ Heterozygous = __________ = _____ = ______ = _________

  10. In a population of pigs color is determined by one gene. If the black allele (b) is recessive and the white allele (B) is dominant, what is the frequency of the black allele in this population? p + q = 1 p2 + 2pq + q2 = 1 ALWAYS START WITH HOMOZYGOUS RECESSIVE (bb) = q2 GENOTYPE OF Black pigs = ______White pigs = ______ or ______ BOZEMAN BIOLOGY SOLVING HARDY WEINBERG PROBLEMS

  11. In a population of pigs color is determined by one gene. If the black allele (b) is recessive and the white allele (B) is dominant, what is the frequency of the black allele in this population? p + q = 1 p2 + 2pq + q2 = 1 Black pigs = ______ = _____ = _____ SO bb . . . q2 = ______

  12. p + q = 1 p2 + 2pq + q2 = 1 b = ______ (q) B = ______ (p) bb = ________ (q2)Bb = _______ (2pq)BB = _______ (p2) If q2 = 0.25q = _____ = _______ p + q = 1 OR 1 – q = p p = 1- 0.5 = _____ If p = ____ then p2 = ______ = ______

  13. p + q = 1 p2 + 2pq + q2 = 1 b = ______ (q) b = ______ (p) bb = ________ (q2)Bb = _______ (2pq)BB = _______ (p2) If you know this :can you figure out 2pq? p2 + 2pq + q2 = 1 ____ + 2pq + ____ = 1 2pq = 1 – 0.5 = _____

  14. p + q = 1 p2 + 2pq + q2 = 1 b = ______ (q) B = _____ (p) bb = ________ (q2)Bb = _______ (2pq)BB = _______ (p2) You can answer ?’s nowWhat is the frequency of thewhite allele (B) in population? B = ___ = _____ = ____ What percent of the population is HOMOZYGOUS DOMINANT? BB = ____ = _____ = _____

  15. In a population of fruit flies, 36% have red eyes and the remainder have sepia eyes. The sepia (r) eye trait is recessive to red (R) eyes. Calculate the frequencies in this population. p + q = 1 p2 + 2pq + q2 = 1 ALWAYS START WITH HOMOZYGOUS RECESSIVE (rr) = q2 GENOTYPE OF Red eyes = ______ or _____sepia eyes = ______

  16. In a population of fruit flies, 64 % have red eyes and the remainder have sepia eyes. The sepia (r) eye trait is recessive to red (R) eyes. Calculate the frequencies in this population. p + q = 1 p2 + 2pq + q2 = 1 If red eyes = ______= _____ ? % have sepia eyes = _____ = r r = q2

  17. p + q = 1 p2 + 2pq + q2 = 1 r = ______ (q) R = ______ (p) rr = ________ (q2)Rr = _______ (2pq)RR = _______ (p2) If q2 = 0.36q = _____ = _______ p + q = 1 OR 1 – q = p p = 1- 0.6 = _____ If p = ____ then p2 = ______ = ______

  18. p + q = 1 p2 + 2pq + q2 = 1 r = ______ (q) R = ______ (p) rr = ________ (q2)Rr = ______ (2pq)RR = _______ (p2) If you know this :can you figure out 2pq? p2 + 2pq + q2 = 1 ____ + 2pq + ____ = 1 2pq = 1 – 0.52 = _____

  19. p + q = 1 p2 + 2pq + q2 = 1 r = ______ (q) R = ______ (p) rr = ________ (q2)Rr = _______ (2pq)RR = _______ (p2) You can answer ?’s nowWhat is the frequency of thesepia allele (r) in population? r = ___ = _____ = ____ What percent of the population is HETEROZYGOUS? Rr = ____ = _____ = _____

  20. BB Bb bb Using Hardy-Weinberg equation p2=.36 2pq=.48 q2=.16 Assuming H-W equilibrium BB Bb bb p2=.20 p2=.74 2pq=.64 2pq=.10 q2=.16 q2=.16 Sampled data Change in allele frequency = EVOLUTION happened

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