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Chpt. 23 The Evolution of Populations-- Population Genetics.

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Presentation on theme: "Chpt. 23 The Evolution of Populations-- Population Genetics."— Presentation transcript:

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2 Chpt. 23 The Evolution of Populations-- Population Genetics

3 Individuals are selected… Populations evolve

4 Populations evolve

5 Populations Populations = unit of evolution Natural selection Natural selection = mechanism of evolution Gradualism Gradualism = accumulation of small changes in gene poolover LONG periods of time

6  Hardy-Weinberg Theorem

7  examines the gene structure of a NON-evolving population

8  Hardy-Weinberg Theorem  Hardy-Weinberg Theorem examines the gene structure of a NON-evolving population.  Obviously, this is not common, however, gives a base-line / model NULL HYPOTHESIS for determining if and why populations evolve

9  Hardy-Weinberg Theorem  Hardy-Weinberg Theorem even though alleles are shuffled and recombined during meiosis and random fertilization. This has no effect on the overall gene pool percentages.

10 Not SWIMMING pool…. A a A a A a A a a a A A A a GENE POOL!!!

11  Hardy-Weinberg Theorem Gene pool frequencies (percentages) will remain unchanged if no mechanism that can cause evolution to occur acts on a population.

12  Hardy-Weinberg Theorem Gene pool frequencies will remain unchanged if: Mutations are not occurring

13  Hardy-Weinberg Theorem Gene pool frequencies will remain unchanged if: Natural selection is not occurring

14  Hardy-Weinberg Theorem Gene pool frequencies will remain unchanged if: Population is LARGE

15  Hardy-Weinberg Theorem Gene pool frequencies will remain unchanged if: EVERYONE breeds…

16  Hardy-Weinberg Theorem Gene pool frequencies will remain unchanged if: EVERYONE randomly mates…

17  Hardy-Weinberg Theorem Gene pool frequencies will remain unchanged if: EVERYONE produces the same number of offspring

18  Hardy-Weinberg Theorem Gene pool frequencies will remain unchanged if: NOONE migrates in or out of the population… everyone stays

19  Hardy-Weinberg Theorem

20 5 Agents of evolutionary change: MutationGene Flow (migration) Genetic Drift (same # of offspring) Selection Non-random mating

21  Hardy-Weinberg Theorem remember: H.W. explains: the frequency of alleles remains constant in a population… unless acted upon by agents OTHER THAN sexual recombination.

22 Hardy-Weinberg Principle p2 + 2pq + q2 = 1 Mathematical statement about the relative frequency of alleles (genotypes) in a population. p +q = 1

23  Hardy-Weinberg Theorem  Frequency of alleles remains constant in a population, unless acted upon by agents OTHER THAN sexual recombination.  Inheritance does not cause changes in allele frequency.

24 Hardy-Weinberg Principle p + q = 1 p = frequency of dominant allele

25 Hardy-Weinberg Principle p + q = 1 q = frequency of recessive allele

26 Hardy-Weinberg Principle in most cases, we only know the phenotypic frequencies Mathematical statement about the relative frequency of alleles (genotypes) in a population.

27 Hardy-Weinberg Principle q 2 = # of aa individuals q 2 = frequency of homozygote recessive individuals

28 Hardy-Weinberg Principle p 2 = # of AA individuals p 2 = frequency of homozygote dominant individuals

29 Hardy-Weinberg Principle p = # of (AA) + 2 (# Aa) p = frequency of dominant allele

30 Hardy-Weinberg Principle q = # of (aa) + 2 (# Aa) q = frequency of recessive allele

31 Hardy-Weinberg Principle 2pq = # of Aa individuals 2pq = frequency of heterozygote individuals

32 Hardy-Weinberg Principle by comparing genotypic frequencies from one generation to the next, you can learn whether or not evolution has occurred…

33 Hardy-Weinberg Principle if genotypic frequencies have changed from your original count… evolution has occurred!

34 Suppose there are 1,000 individuals in a population Genotype NumberGenotypic Frequency AA 4900.49 Aa 4200.42 aa 90 0.09 total 10001.00

35 Suppose there are 1,000 individuals in a population Genotypic Frequency 0.49 0.42 0.09 total1.00 Genotypic frequency = the proportion of a particular genotype found in a population AA Aa aa

36 Suppose there are 1,000 individuals in a population Phenotype NumberPhenotypic Frequency dominant 9100.91 recessive 900.09 total 10001.00

37 Suppose there are 1,000 individuals in a population Phenotypic Frequency 0.91 0.09 total1.00 Phenotypic frequency = the proportion of a particular phenotype found in a population

38 Suppose there are 1,000 individuals in a population Allele NumberAllele Frequency A 14000.7 a 4200.3 total 20001.00

39 480 Allele frequency Pssst…(There are 1,000 copies of the flower color gene in this population of 500 total flowers…) q = q = frequency of recessive allele However, we do not know how many a’s there are just by looking at phenotype

40 480 Genotypic frequency q 2 = 20/500 q2 = frequency of recessive genotype

41 480 Genotypic frequency q 2 =.04 q2 = frequency of recessive genotype

42 480 Allele frequency q = .04 q = frequency of recessive allele

43 480 Allele frequency q =.2 q = frequency of recessive allele

44 480 Allele frequency q =.2 q = frequency of recessive allele p = frequency of dominant allele p + q = 1 p +.2 = 1 p = 1 -.2 p =.8

45 480 Some of the pink flowers will be AA and some will be Aa p2 + 2pq + q2 = 1 p =.8 q =.2

46 480 Some of the pink flowers will be AA and some will be Aa.64 +.32 +.04 = 1

47 480 How many of the pink flowers will be AA and how many will be Aa.64 X 500 individuals 320 individuals are AA

48 480 How many of the pink flowers will be AA and how many will be Aa.32 X 500 individuals 160 individuals are Aa

49 480 320 are AA 320 are AA 160 are Aa 160 are Aa 480 total

50 Genetic structure of next generation.8 x.8 =.64

51 Genetic structure of next generation.2 x.2 =.04

52 Genetic structure of next generation.2 x.8 =.16.8 x.2 =.16.32 aA Aa

53 Hardy-Weinberg Principle Under ideal conditions, the relative allele frequencies are the same in the offspring generation as in the parent generation. p 2 + 2pq + q 2 = 1 p 2 + 2pq + q 2 = 1

54 IF IF: frequencies in a population deviate from Hardy-Weinberg (these are set numbers)… THEN THEN: we know the population is evolving.

55 Oh Hardy, why did you quit your job at ABERCRO MBIE? Weinburg, I keep telling you, I got sick of changing genes!!!

56 change over time is a result of changes in a population’s frequency of genotypes / genetic

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59 Lets Hardy!! What percentage of the human population are carriers for the allele for PKU? 1 out of 10,000 babies are born with this recessive disease Most of the time you will begin by determining the frequency of the homozygous recessive genotype Find q2 (frequency of homo recessive)

60 Lets Hardy!! What percentage of the human population are carriers for the allele for PKU? 1 out of 10,000 babies are born with this recessive disease Hint: q2 = 1 / 10,000 (frequency of aa)  Find q  Find p  Find pq

61 AP Problems Using Hardy-Weinberg Solve for q 2 (% of total) Solve for q (equation) Solve for p (1- q) H-W is always on the national AP Bio exam (but no calculators are allowed).

62 AP Problems Using Hardy-Weinberg Solve for q 2 (% of total) Solve for q (equation) Solve for p (1- q) H-W is always on the national AP Bio exam (but no calculators are allowed).

63 AP Problems Using Hardy-Weinberg Solve for q 2 (% of total) Solve for q (equation) Solve for p (1- q) H-W is always on the national AP Bio exam (but no calculators are allowed).

64 AP Problems Using Hardy-Weinberg Solve for q 2 (% of total) Solve for q (equation) Solve for p (1- q) H-W is always on the national AP Bio exam (but no calculators are allowed)

65 AP Problems Using Hardy-Weinberg population: 100 cats 84 black, 16 white How many of each genotype? population: 100 cats 84 black, 16 white How many of each genotype? q 2 (bb): 16/100 =.16 0.4 q (b): √.16 = 0.4 0.6 p (B): 1 - 0.4 = 0.6 q 2 (bb): 16/100 =.16 0.4 q (b): √.16 = 0.4 0.6 p (B): 1 - 0.4 = 0.6 bbBbBB p 2 =.36 2pq=.48 q 2 =.16

66 AP Problems Using Hardy-Weinberg bbBbBB p 2 =.36 2pq=.48 q 2 =.16 Assuming H-W equilibrium Sampled data bbBbBB p 2 =.20 2pq=.64 q 2 =.16 How do you explain the data? Null hypothesis


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