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2
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- Population genetics is the study of allele behaviour in populations.
- A population is a group of interbreeding individuals of a single
species.
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3
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- Variation within a population
- Discrete characteristics – Purple vs. White flower colours
- Quantitative characters – (two or more genes) vary along a continuum
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4
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- Nonheritable variation
- Nemoria arizonaria moths
- Fed oak flowers Fed oak leaves
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- All new genes arise by mutation (point mutation) creates new codons and
therefore new protein sequences
- Sexual reproduction provides recombination (meiosis and crossing over)
which produces endless genotypic variation in the population
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7
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- This theorem describes a population that is not evolving.
- That is the populations allelic frequency is not changing
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9
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10
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- This population of “circle critters” has twenty members, and each member
has two alleles, therefore there 40 alleles in the “gene pool”
- The allelic frequency of A is 36/40=0.9
- The allelic frequency of a is 4/40=0.1
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- Therefore
- 90% of all the sperm will carry A
- 10% of all the sperm will carry a
- 90% of all the eggs will carry A
- 10% of all the eggs will carry a
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12
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13
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14
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15
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16
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17
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18
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- If p = frequency of allele A = 0.9
- And q = frequency of allele a = 0.1
- Then AA + 2Aa + aa = 1 would be
- p2 + 2pq + q2 = 1
- From the above example p = 0.9, and
- q = 0.1
- (0.9)2 + 2(0.9)(0.1) + (0.1)2 = 1
- 0.81 + 0.18 + 0.01 = 1
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- The term p2 = frequency of genotype AA = 0.81
- The term 2pq = frequency of genotype Aa = 0.18
- The term q2 = frequency of genotype aa = 0.01
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- Lets say aa represents blond hair,
- and 4% of a population have blond hair.
- Therefore aa = 0.04
- We know that q2 = the frequency of aa
- Therefore q =
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- Lets say aa represents blond hair,
- and 4% of a population have blond hair.
- Therefore aa = 0.04
- We know that q2 = the frequency of aa
- Therefore q = 0.04 = 0.2
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- Lets say aa represents blond hair,
- and 4% of a population have blond hair.
- Therefore aa = 0.04
- We know that q2 = the frequency of aa
- Therefore q = 0.04 = 0.2
- Recall that p + q =1
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- Lets say aa represents blond hair,
- and 4% of a population have blond hair.
- Therefore aa = 0.04
- We know that q2 = the frequency of aa
- Therefore q = 0.04 = 0.2
- Recall that p + q =1
- If q = 0.2, then 1 - 0.2 = p = 0.8
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- p2 + 2pq + q2 = 1
- (0.8)2 + 2(0.8)(0.2) + (0.2)2 = 1
- AA = 0.64
- Aa = 0.32
- aa = 0.04
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- The population must be large
- Mutations must not occur
- There must be no immigration or emigration
- Reproduction must be random
- There must be no natural selection
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30
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31
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32
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33
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34
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- Directional Selection
The British Peppared Moth
Biston betularia
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36
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37
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38
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39
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40
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- Selection “within the same sex”.
Male compete with each other for females.
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41
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- Also called “mate choice”. Often
female choose the males.
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42
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43
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44
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- “Changing the locks” – bacterial pathogens and the antigens they bind
to.
- “Red Queen race” – Co-evolution of host parasite interactions.
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45
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- Diploidy – genetic variation hidden from selection by recessive alleles.
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46
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- Heterozygotic Advantage- Sickle Cell Anemia
- Hemoglobin Hs
- Hemoglobin Ha
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47
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- Hemoglobin Hs
- Hemoglobin Ha
- If Ha was 0.8 =p and Hs was 0.2 = q
- Then from the Hardy-Weinberg equation
- p2 + 2pq
+ q2 = 1
- Ha Ha Ha
Hs Hs Hs
- 0.64 + 0.32 + 0.04 = 1
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48
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- But before the next generation Hs Hs would
- likely die and not be available to reproduce.
- Therefore 0.64 = Ha Ha
- 0.32 = Ha Hs
- Total 0.96
- Now 0.64/0.96 = 0.66 = Ha Ha
- and 0.32/0.96 = 0.33 = Ha Hs
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49
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50
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51
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52
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- The fitness of a phenotype declines if it becomes too common in the
population.
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53
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- Neutral Variation
- Not all traits have an impact on reproductive success
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