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Hardy-Weinberg Conditions

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Random Mating (especially affects genotype frequencies) ... If the alleles are B and b (for blue or white flowers), how many B alleles are ... – PowerPoint PPT presentation

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Title: Hardy-Weinberg Conditions


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  • Hardy-Weinberg Conditions
  • No Net Mutations
  • No Net Migrations
  • Random Mating
    (especially affects genotype frequencies)
  • Large Population Size
    (limits effects of drift)
  • No Selection
    (all phenotypes of equal
    fitness)

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No Net Mutations
new dom. alleles ( dom. alleles) (
forward mutations) ( reverse mutations)
p frequency of dominant allele q frequency
of recessive allele
u frequency of mutation from dominant allele to
recessive (forward) v frequency of mutation
from recessive allele to dominant (reverse)
Practice Example population of flowers where
n100,000 and p0.6 u10-3 and v10-4 (rates are
changes/allele)
If the alleles are B and b (for blue or white
flowers), how many B alleles are present before
any mutations?
0.6 x 200,000 120,000
How many B alleles are present after mutations in
that generation?
120,000 (0.001 x 120,000) (0.0001 x 80,000)
120,000 120 8 119,888
Assuming there are still 100,000 flowers, what is
the new p?
119,888/200,000 0.599
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No Net Migrations
px m py (1-m)px
px frequency of dominant allele in population
x py frequency of dominant allele in
population y m proportion of study population
that is made up of migrants (this means the
proportion not made up of migrants is 1-m)
Practice Example px 0.7 and py
0.4 Population x originally had 5000 individuals,
1000 individuals migrate in from population y
Practice Example px 0.7 Population x
originally had 5000 individuals, 400 individuals
migrate out of the population, all are homozygous
recessive
What is the px?
px (1-m)px mpx (mpx 0)
px mpy (1-m)px px 0.4(1000/6000)
0.7(5000/6000) 0.65
(1-m) 5000/4600 (the old px represents a
greater proportion of the current population than
it did in the old population) px 0.7 x 1.087
0.76
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Non-Random Mating
Assortative Mating like mates with like (e.g.,
inbreeding) Disassortative Mating like mates
with unlike (e.g., sweat study) Non-random
mating doesnt alter allele frequencies, but does
alter genotype frequencies
If we start with p0.5 and q0.5 with typical H-W
genotype frequencies, then we have
p and q dont change, only genotypes
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Non-Random Mating
Assortative
Disassortative
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Large Population Size
Small populations have a greater chance of
genetic drift random fluctuations
in allele frequencies over time
Bottlenecks Founder Effects are types of drift
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No Selection
  • Natural Selection requires three things
  • Phenotypic variation among individuals
  • Genetic basis to the phenotypic variation
  • Differential reproductive success among phenotypes

Differential survival and reproductive success
fitness (W) Fitness is all about of genes
passed on Genotype(s) that survive the most have
a fitness of 1 Fitness for other genotypes is
expressed as the proportion of survival relative
to the most fit
Selection coefficient (s) is the opposite of
fitness (1-W) If genotypes TT and Tt produce the
most offspring and genotype tt produces 20
fewer, then Wtt 0.8 and stt 0.2
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No Selection
Practice Example A trait exhibiting simple
dominance exists in a population. The dominant
phenotype (from genotypes AA and Aa) is the most
fit. For every 100 offspring produced by dominant
phenotype individuals, only 78 are produced from
recessive phenotype individuals.
The dominant allele is not always the most
fit. If there is a modified dominance
interaction, the heterozygous genotype could be
the most fit heterosis.
What are the fitnesses of the three genotypes?
WAA 1, WAa 1, Waa .78
What are the selection coefficients of the three
genotypes?
sAA 0, sAa 0, saa .22
What are the expected genotypic contributions to
the next generation if the current frequencies
are p2 0.25, 2pq 0.5, q2 0.25?
genotype contribution current frequency x
fitness WAAp2 1 x 0.25 WAa2pq 1 x 0.5 Waaq2
0.78 x 0.25
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WAA 1, WAa 1, Waa .78
sAA 0, sAa 0, saa .22
p2 0.25, 2pq 0.5, q2 0.25
WAAp2 0.25, WAa2pq 0.5, Waaq2 0.195
If the current population contains 1000
individuals, and the next generation contains
1000 individuals, how many in each are aa?
aa in original 1000 x 0.25 250
aa in new ((0.25 x 0.78)/0.945) x 1000 206
a in new? (206 x 2) (((0.5 x 1)/0.945) x
1000) 941
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  • Population genetics used to answer ecological
    questions
  • Is there selection occurring? fitness of
    individuals? population?
  • What are the mating systems like? assortative?
    disassortative?
  • inbreeding?
  • What kind of genetic variation is available for
    adaptation?
  • What kind of gene flow is occurring?
  • Conservation questions what species/populations
    should be conserved? Has introgression occurred?
    is this important?
  • MVP PVA
  • Identifying genetics in populations
  • Visible Polymorphisms
  • Molecular Markers
  • Proteins (allozymes)
  • RFLPs, VNTRs, STRs, etc.
  • sequencing

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