Brachydactyly and evolutionary change

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Brachydactyly and evolutionary change

• We know the gene for brachydactyl fingers is
  dominant to normal fingers
• A man named Yule suggested that short-fingered
  people should become more common through time
• Godfrey Hardy showed this inference was wrong
• Wilhelm Weinberg derived the same solution to
  the problem independently

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The Hardy-Weinberg Law - the most important
principle in population genetics

• The law is divided into three parts a set of
  assumptions and two major results
• In an infinitely large, randomly mating
  population, free from mutation, migration and
  natural selection (note there are five
  assumptions here)
• the frequencies of the alleles do not change and
• as long as the mating is random, the genotypic
  frequencies will remain in the proportions p2
  (frequency of AA), 2pq (frequency of Aa) and q2
  (frequency of aa) where p is the frequency of A
  and q is the frequency of a
• The sum of the genotypic frequencies should be
  p2 2pq q2 1

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Figure 22.3
DERIVING THE HARDY-WEINBERG PRINCIPLE-A NUMERICAL
EXAMPLE
P1 frequency of allele A1 0.7
1. Suppose that the allele frequencies in the
parental generation were 0.7 and 0.3.
P2 frequency of allele A2 0.3
Gametesfrom parent generation
2. 70 of the gametes in the gene pool carry
allele A1 and 30 carry allele A2 .
3. Pick two gametes at random from the gene pool
to form offspring. Three genotypes are possible.
A2
A1
A1
A2
A1
A1
A2
A2
.07 x 0.30.21
.03 x 0.70.21
0.7 x 0.7 0.49
0.3 x 0.3 0.09
0.21 0.21 0.42
Homozygous
Heterozygous
Homozygous
4. Calculate the frequencies of these three
combinations of alleles.
Gametesfrom offspring generation
5. When the offspring breed, imagine that their
gametes go into a gene pool.
6. Calculate the frequencies of the two alleles
in this gene pool.
42 of the gametes are from A1A2 parents. Half of
these carry A1and half carry A2
49 of the gametes are from A1A1 parents. All of
these carry A1
9 of the gametes are from A2A2 parents. All of
these carry A2
BEHOLD! The allele frequencies of A1and A2 have
not changed from parent generation to offspring
generation. Evolution has not occurred.
P1 frequency of allele A1 (0.49 1/2(0.42))
(0.49 0.21) 0.7
P2 frequency of allele A2 (1/2(0.42) 0.09)
(0.21 0.09) 0.3
Genotype frequencies will be given by p12
2p1p2 p22 as long as all Hardy-Weinberg
assumptions are met
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Genotypic frequencies under the Hardy-Weinberg
Law

• The Hardy-Weinberg Law indicates
• At equilibrium, genotypic frequencies depend on
  the frequencies of the alleles
• The maximum frequency for heterozygotes is 0.5
• If allelic frequencies are between 0.33 and 0.66,
  the heterozygote is the most common genotype

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Albinism in Hopi Native Americans

• The incidence of albinism is remarkably common
  (0.0043 or 13 in every 3000 Hopis)
• Assuming Hardy-Weinberg equilibrium, we can
  calculate q as the square root of 0.0043 0.066
• p is therefore equal to 0.934
• The frequency of heterozygotes in the population
  is 2pq 0.123
• In other words, 1 in 8 Hopis carries the gene for
  albinism!
• Take-home Lesson For a rare allele,
  heterozygotes can be relatively common

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Extensions of the Hardy-Weinberg Law

• X-linked traits (two alleles)
• Females have two copies of the gene but males are
  hemizygous
• Female genotypes should be in the familiar p2
  2pq q2 ratio
• Male genotypes should be in a pq ratio
• This result shows why X-linked recessive traits
  are much more frequently observed in males

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Extensions of the Hardy-Weinberg Law

• Multiple alleles
• Let p, q and r represent the frequencies of the
  three alleles
• p q r 1.0
• The Hardy-Weinberg model predicts the genotype
  frequencies will be (p q r)2
• The result is a ratio of p2 2pq q2 2pr r2 2pq

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Testing for Hardy-Weinberg distributions

• If we know the frequency of genotypes in a
  population, we can test to see if those
  frequencies conform to the expectations from the
  Hardy-Weinberg model
• Lets reconsider the data for scarlet tiger moths
• Number of BB 452
• Number of Bb 43
• Number of bb 2

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• Now we need to determine the allele frequency of
  B and b
• Lets determine the frequency of allele B
• p f(B) (2 x 452) 43/994 0.953
• We can determine the value of q by difference
• q 1 - p 1 - 0.953 0.047
• Next we determine the expected Hardy-Weinberg
  frequencies of genotypes (p22pq q2)
• That ratio is 0.908 BB 0.090 Bb 0.002 bb

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• Now we can apportion the 497 individuals to their
  expected genotypes
• Expected BB 0.908 x 497 451
• Expected Bb 0.090 x 497 45
• Expected bb 0.002 x 497 1
• Compare the observed genotypes in the field 453,
  42 and 2
• These moths clearly conform to the Hardy-Weinberg
  expectations