Selection and Mutation

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Chapter 6

• Selection and Mutation

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Adding Selection to the Hardy-Weinberg Analysis

• If either of the following occurs then the
  population is responding to selection.
• 1. Some phenotypes allow greater survival to
  reproductive age.
• -or-
• 2. All individuals reach reproductive age but
  some individuals are able to produce more viable
  (reproductively successful) offspring.
• If these differences are heritable then
  evolution may occur over time.

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Caution

• It needs to be mentioned that most phenotypes are
  not strictly the result of their genotypes.
• Environmental plasticity and
• interaction with other genes may also be
  involved.
• In other words it is not as simple as we are
  making it here but we have to start somewhere.

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We will look at two possible effects of natural
selection on the gene pool.

• Selection may alter allele frequencies or violate
  conclusion 1
• Selection may upset the relationship between
  allele frequencies and genotype frequencies.
• Conclusion 1 is not violated but conclusion 2
  is violated.
• In other words the allele frequencies remain
  stable but genotype frequencies change and can no
  longer be predicted accurately from allele
  frequencies.

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FIRST An example of what we might normally see
happen to allele frequencies when natural
selection is at work
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Let B1 and B2 the allele frequencies of the
initial population with frequencies of B1 .6
and B2 .4

• After random mating which produces 1000 zygotes
  we get

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Selection Example
Initial frequencies B1 0.6 B2 0.4 B1B1 B1B2 B2B2 1000 total






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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 B1B2 B2B2 1000 total






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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 B2B2 1000 total






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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total






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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes





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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes 100 survive





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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes 100 survive 75 survive





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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes 100 survive 75 survive 50 survive





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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes 100 survive 75 survive 50 survive
number surviving




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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes 100 survive 75 survive 50 survive
number surviving 360




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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes 100 survive 75 survive 50 survive
number surviving 360 360




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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes 100 survive 75 survive 50 survive
number surviving 360 360 80




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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes 100 survive 75 survive 50 survive
number surviving 360 360 80 800 total




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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes 100 survive 75 survive 50 survive
number surviving 360 360 80 800 total
The genotype frequencies of mating individuals which survive is



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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes 100 survive 75 survive 50 survive
number surviving 360 360 80 800 total
The genotype frequencies of mating individuals which survive is .45



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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes 100 survive 75 survive 50 survive
number surviving 360 360 80 800 total
The genotype frequencies of mating individuals which survive is .45 .45



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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes 100 survive 75 survive 50 survive
number surviving 360 360 80 800 total
The genotype frequencies of mating individuals which survive is .45 .45 .10



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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes 100 survive 75 survive 50 survive
number surviving 360 360 80 800 total
The genotype frequencies of mating individuals which survive is .45 .45 .10
The resulting allelic frequencies in the new reproducing population is B1 B2


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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes 100 survive 75 survive 50 survive
number surviving 360 360 80 800 total
The genotype frequencies of mating individuals which survive is .45 .45 .10
The resulting allelic frequencies in the new reproducing population is B1 .451/2(.45) 0.675


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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes 100 survive 75 survive 50 survive
number surviving 360 360 80 800 total
The genotype frequencies of mating individuals which survive is .45 .45 .10
The resulting allelic frequencies in the new reproducing population is B1 .451/2(.45) 0.675 B2


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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes 100 survive 75 survive 50 survive
number surviving 360 360 80 800 total
The genotype frequencies of mating individuals which survive is .45 .45 .10
The resulting allelic frequencies in the new reproducing population is B1 .451/2(.45) 0.675 B2 1/2(.45)0.10 0.325


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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes 100 survive 75 survive 50 survive
number surviving 360 360 80 800 total
The genotype frequencies of mating individuals which survive is .45 .45 .10
The resulting allelic frequencies in the new reproducing population is B1 .451/2(.45) 0.675 B2 1/2(.45)0.10 0.325
an increase of .075 a decrease of .075

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Selection Example
Initial frequencies B1 0.6 B2 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total
differential survival of offspring leads to reduced numbers of some genotypes 100 survive 75 survive 50 survive
number surviving 360 360 80 800 total
The genotype frequencies of mating individuals which survive is .45 .45 .10
The resulting allelic frequencies in the new reproducing population is B1 .451/2(.45) 0.675 B2 1/2(.45)0.10 0.325
an increase of .075 a decrease of .075
Thus, conclusion 1 is violated and the allele frequencies are changing we are not in equilibrium. The population is evolving! Thus, conclusion 1 is violated and the allele frequencies are changing we are not in equilibrium. The population is evolving! Thus, conclusion 1 is violated and the allele frequencies are changing we are not in equilibrium. The population is evolving! Thus, conclusion 1 is violated and the allele frequencies are changing we are not in equilibrium. The population is evolving!
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Sample analysis of a population for the effects
of selection

• analyze the population on the basis of the
  fitness of the offspring produced.
• The fittest individuals will survive the
  selection process and leave offspring of their
  own.
• We are going to define fitness as the survival
  rates of individuals which survive to reproduce.

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Fitness formulas
MEAN FITNESS

• If
• w11 fitness of allele 1 homozygote (exp
  B1B1)
• w12 fitness of the heterozygote (exp B1B2)
• w22 fitness of allele 2 homozygote exp
  (B2B2)

mean fitness of the population will be described
by the formula
w p2w11 2pqw12 q2w22
CAUTION! Use ONLY allele frequencies in these
formulas NOT genotype frequencies!
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For our previous example

• B1 0.6 and B2 0.4 and
• fitness of B1B1 1.0 (100 survived)
• fitness of B1B2 .75 ( 75 survived)
• fitness of B2B2 .50 (50 survived)
• Figure the mean fitness now.

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For our previous example

• B1 0.6 and B2 0.4 and
• fitness of B1B1 1.0 (100 survived)
• fitness of B1B2 .75 ( 75 survived)
• fitness of B2B2 .50 (50 survived)
• Figure the mean fitness now.
• w (.6)2(1)

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For our previous example

• B1 0.6 and B2 0.4 and
• fitness of B1B1 1.0 (100 survived)
• fitness of B1B2 .75 ( 75 survived)
• fitness of B2B2 .50 (50 survived)
• Figure the mean fitness now.
• w (.6)2(1)(2(.6)(.4)(.75))

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For our previous example

• B1 0.6 and B2 0.4 and
• fitness of B1B1 1.0 (100 survived)
• fitness of B1B2 .75 ( 75 survived)
• fitness of B2B2 .50 (50 survived)
• Figure the mean fitness now.
• w (.6)2(1)(2(.6)(.4)(.75)) (.4)2 (.5) .80

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We can also use the fitness to calculate the
expected frequency of each genotype in the next
generation. We can do it the long way , if we
know actual numbers OR.
B1B1 P2w11 w B1B2
2pqw12 w B2B2 q2w22
w
We can use these formulas which can calculate
the new expected genotype frequencies based on
the fitness of each genotype and the allele
frequencies in the current generation.
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We can also use the fitness values and current
allele frequencies to calculate the expected
frequency of each allele in the next generation.
B1 p2w11pqw12 B2 pqw12q2w22
w w
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Finally, we can calculate the change (?) in the
frequency of alleles B1 or B2 directly as
follows
? B1 ?p p (pw11qw12 w)
w
? B2 ?q q (pw12qw22 w)
w
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Lets do a sample problem.

• Go back to the problem we did in class last time.
  Taking this current population that you have
  already analyzed, figure out what the new
  genotype and allele frequencies will be if the
  fitness of these individuals is actually as
  follows
• SS individuals 0.8 Ss individuals 1.0 and the
  ss individuals 0.6.

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Last time we calculated S .82 and s .18
Now we set the fitnesses at w11(SS).8w12(Ss)1
w22(ss).6
Calculate the w and B1B1 B1B2 and B2B2 values
for the next generation now
w p2w11 2pqw12 q2w22
w (.82) 2 (.8) 2(.82)(.18)(1.0) (.18)2 (.6)
w .537 .295 .019 .85
B1B1 P2w11 w B1B2
2pqw12 w B2B2 q2w22
w
SS (.82)2(.8) / .85 .633
Ss 2(.82)(.18)(1.0) / .85 .347
ss (.18)2(.6) / .85 0.023
Hint Do they add up to 1.0?
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We an also calculate the new allele frequencies
as well
S (.82)2(.8) (.82)(.18)(1.0) .806
.85
B1 p2w11pqw12 w
s (.82)(.18)(1.0) (.18)2(.6) .196
.85
B2 pqw12q2w22 w
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So B1B1 .63 B1B2 .35
B2B2 .02
and
B1 .80 B2 .20
Is this population in equilibrium?
Have the allele frequencies changed?
Can we predict the genotype frequencies from the
allelic frequencies?
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Experimental confirmation of loss of Hardy
Weinberg equilibrium
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• Fruit fly experiments of
• Cavener and Clegg

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• Worked with fruit flies having two versions of
  the ADH (alcohol dehydrogenase) enzyme, F and S.
  (for fast and slow moving through an
  electrophoresis gel)
• Grew two experimental populations on food spiked
  with ethanol and two control populations on
  normal, non-spiked food. Breeders for each
  generation were picked at random.
• Took random samples of flies every few
  generations and calculated the allele frequencies
  for AdhF and AdhS

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Figure 6.13 pg 185
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Can we identify which assumption is being
violated?

• only difference is ethanol in food
• no migration
• assured random mating
• population size, drift?
• no mutation.
• Must be selection for the fast form of gene.
• Indeed studies show that AdhF form breaks down
  alcohol at twice the rate as the AdhS form.
• Therefore offspring carrying this allele are more
  fit and leave more offspring and the make-up of
  gene pool changes.

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Cavener and Clegg demonstrated that selection
pressure can lead to changes in allele
frequencies in just a few generations
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A second selection scenario

• Selection may upset the relationship between
  allele frequencies and genotype frequencies.
• Conclusion 1 ( allele frequencies do not change)
  is not violated but conclusion 2 (that we can
  predict genotype frequencies from allele
  frequencies) is violated.

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A sample calculation
Initial B1 0.5 Initial B2 0.5 250 B1B1 500 B1B2 250 B2B2 1000 Total
Differential fitness of the genotypes Fitness .50 Fitness 1.0 Fitness .50
Number of survivors to reproductive age 125 500 125 750 total





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A sample calculation
Initial B1 0.5 Initial B2 0.5 250 B1B1 500 B1B2 250 B2B2 1000 Total
Differential fitness of the genotypes Fitness .50 Fitness 1.0 Fitness .50
Number of survivors to reproductive age 125 500 125 750 total
The genotype frequencies of mating individuals which survive




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A sample calculation
Initial B1 0.5 Initial B2 0.5 250 B1B1 500 B1B2 250 B2B2 1000 Total
Differential fitness of the genotypes Fitness .50 Fitness 1.0 Fitness .50
Number of survivors to reproductive age 125 500 125 750 total
The genotype frequencies of mating individuals which survive 125 / 750 0.167




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A sample calculation
Initial B1 0.5 Initial B2 0.5 250 B1B1 500 B1B2 250 B2B2 1000 Total
Differential fitness of the genotypes Fitness .50 Fitness 1.0 Fitness .50
Number of survivors to reproductive age 125 500 125 750 total
The genotype frequencies of mating individuals which survive 125 / 750 0.167 500 / 750 0.667




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A sample calculation
Initial B1 0.5 Initial B2 0.5 250 B1B1 500 B1B2 250 B2B2 1000 Total
Differential fitness of the genotypes Fitness .50 Fitness 1.0 Fitness .50
Number of survivors to reproductive age 125 500 125 750 total
The genotype frequencies of mating individuals which survive 125 / 750 0.167 500 / 750 0.667 125 / 750 0.167
The resulting allelic frequencies in new mating population



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A sample calculation
Initial B1 0.5 Initial B2 0.5 250 B1B1 500 B1B2 250 B2B2 1000 Total
Differential fitness of the genotypes Fitness .50 Fitness 1.0 Fitness .50
Number of survivors to reproductive age 125 500 125 750 total
The genotype frequencies of mating individuals which survive 125 / 750 0.167 500 / 750 0.667 125 / 750 0.167
The resulting allelic frequencies in new mating population B1 .1671/2 (0.667) 0.5



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A sample calculation
Initial B1 0.5 Initial B2 0.5 250 B1B1 500 B1B2 250 B2B2 1000 Total
Differential fitness of the genotypes Fitness .50 Fitness 1.0 Fitness .50
Number of survivors to reproductive age 125 500 125 750 total
The genotype frequencies of mating individuals which survive 125 / 750 0.167 500 / 750 0.667 125 / 750 0.167
The resulting allelic frequencies in new mating population B1 .1671/2 (0.667) 0.5 B2 ½ (.667) .167 0.5



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A sample calculation
Initial B1 0.5 Initial B2 0.5 250 B1B1 500 B1B2 250 B2B2 1000 Total
Differential fitness of the genotypes Fitness .50 Fitness 1.0 Fitness .50
Number of survivors to reproductive age 125 500 125 750 total
The genotype frequencies of mating individuals which survive 125 / 750 0.167 500 / 750 0.667 125 / 750 0.167
The resulting allelic frequencies in new mating population B1 .1671/2 (0.667) 0.5 B2 ½ (.667) .167 0.5
No change No change


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A sample calculation
Initial B1 0.5 Initial B2 0.5 250 B1B1 500 B1B2 250 B2B2 1000 Total
Differential fitness of the genotypes Fitness .50 Fitness 1.0 Fitness .50
Number of survivors to reproductive age 125 500 125 750 total
The genotype frequencies of mating individuals which survive 125 / 750 0.167 500 / 750 0.667 125 / 750 0.167
The resulting allelic frequencies in new mating population B1 .1671/2 (0.667) 0.5 B2 ½ (.667) .167 0.5
No change No change
Thus conclusion 1 is not violated therefore this population has not evolved..but.. Thus conclusion 1 is not violated therefore this population has not evolved..but.. Thus conclusion 1 is not violated therefore this population has not evolved..but.. Thus conclusion 1 is not violated therefore this population has not evolved..but..

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A sample calculation
Initial B1 0.5 Initial B2 0.5 250 B1B1 500 B1B2 250 B2B2 1000 Total
Differential fitness of the genotypes Fitness .50 Fitness 1.0 Fitness .50
Number of survivors to reproductive age 125 500 125 750 total
The genotype frequencies of mating individuals which survive 125 / 750 0.167 500 / 750 0.667 125 / 750 0.167
The resulting allelic frequencies in new mating population B1 .1671/2 (0.667) 0.5 B2 ½ (.667) .167 0.5
Change in allele frequency No change No change
Thus conclusion 1 is not violated therefore this population has not evolved.but.. Thus conclusion 1 is not violated therefore this population has not evolved.but.. Thus conclusion 1 is not violated therefore this population has not evolved.but.. Thus conclusion 1 is not violated therefore this population has not evolved.but..
Conclusion 2 is violated. We are not in equilibrium. Frequency of B1B1 .167 which is not equal to (.5)2 Conclusion 2 is violated. We are not in equilibrium. Frequency of B1B1 .167 which is not equal to (.5)2 Conclusion 2 is violated. We are not in equilibrium. Frequency of B1B1 .167 which is not equal to (.5)2 Conclusion 2 is violated. We are not in equilibrium. Frequency of B1B1 .167 which is not equal to (.5)2
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Empirical research on Selection and Genotype
frequency

• Kuru example among the Foré in New Guinea
• Pg 188-191
• Wanted to determine if there was a genetic basis
  to the resistance of kuru infection.
• Ritualistic mortuary feasts, only young women ate
  the contaminated nervous system tissue leading to
  CJD (similar to mad cow disease)
• Among young women who never participated he Met
  allele 0.48 and the Val allele 0.52 Genotypes
  were Met/Met 0.22 Met/Val 0.51 and Val/ Val
  0.26 very close to the values expected for H-W.

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Compare to older women who participated in the
feasts but did not get sick

• Met 0.52 and Val 0.48
• The expected genotypes are
• Met/Met 0.27 Met/ Val 0.5 and Val/Val 0.23
• The actual were
• Met/Met 0.13 Met/ Val 0.77 and Val/Val 0.10
• Appears homozygotes are susceptible but
  heterozygotes are protected.

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There are TWO conditions necessary for a
favorable allele to increase in a population due
to selection.

• HIV example in book. pg 191
• Two conditions must be met
• 1. Need a high enough frequency of the
  beneficial allele in the population gene pool
• 2. There must be high selection pressure for the
  allele in the same area. In this case a high
  incidence of HIV infection.

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What pattern of allele frequency changes might be
caused by selection

• If selection is acting, does the rate of
  evolution of a particular allele depend on
  whether it is.

dominant or recessive?
heterozygote or homozygote?
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Selection on Recessive and Dominant alleles

• Tribolium Beetle example

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Natural Selection is most potent as an
evolutionary force against recessive alleles when
recessive alleles are common and the dominant
form is relatively rare.

• Dawsons Flour beetle example
• Studied a gene locus that had a wild type ()
  allele and a lethal allele.
• / or /L are normal L/L is lethal.
• Two experimental populations composed of all
  heterozygotes /L
• Therefore started with 0.5 and L 0.5.
• Expected populations to evolve toward lower
  frequency of the L allele.

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Results showed that the recessive lethal did drop
rapidly at first but slowed down over successive
generations.
WHY?
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In each succeeding generation all LL are lost and
makes up a greater proportion of the survivors.

• As you go on there are less and less homozygous
  lethals for selection to act on and the lethal
  allele hides in the heterozygotes

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Summary

• Dawson showed that dominance and allele frequency
  interact to determine the rate of evolution when
  acted on by selection
• If a recessive allele is common evolution is
  rapid because there are a lot of homozygotes that
  express the phenotype for selection to act on.
• If recessive allele is rare, evolution is slow
  because the rare allele is hidden in the
  heterozygotes where selection cannot act.
• His experiments also demonstrated that
• controlled lab situations can accurately predict
  the course of evolution
• populations do what you would expect if selection
  is occurring as predicted by the evolutionary
  theory.

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Selection on Heterozygotes and Homozygotes

• Normally in a recessive/ dominant gene, the
  fitness of the heterozygote will be equal to one
  of the homozygotes
• Also, it is possible for the heterozygotes to
  have a fitness intermediate to the two
  homozygotes.
• Thirdly we may find Heterozygote Superiority or
  Inferiority

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Scenario 1 from Mukai and BurdickFruit fly
experiment

• Studied a gene in which
• Homozygotes for one allele are viable
• Homozygotes for the other allele are not viable
  and are lethal.
• Heterozygotes have a higher fitness than either
  homozygotes

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The experiment

• Started with all heterozygotes to establish a new
  population (each allele .5)
• After several generations equilibrium was
  reached at .79 frequency for the viable allele
• This means that the lethal allele was maintained
  at frequency of 0.21! How could this be?
• Started more populations beginning with
  frequency of .975 of viable allele. Expect the
  population to eliminate all lethal alleles and
  fix the viable allele at 1.0.
• But .....

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The viable allele dropped in frequency and the
same equilibrium around a frequency of .79 was
reached for the viable allele!
Figure 6.18 pg 200
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This is a case of Heterozygote superiority or
overdominance (also called heterosis)

• There is some advantage to the heterozygote
  condition and the heterozygote actually has a
  superior fitness to either homozygote.
• Example in humans is sickle cell anemia
• Leads to the maintenance of genetic diversity
  balanced polymorphism

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Can also have Heterozygote inferiority or
underdominance

• Where the heterozygote condition is inferior to
  either of the homozygotes
• What do you predict would happen to the allele
  frequencies here?
• Leads to fixation of one allele in the
  population, while the other is lost.
• Either allele may be fixed depending on
  conditions and beginning frequencies of each
  allele in the gene pool.

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What is the evolutionary impact?

• Leads to a loss of genetic diversity
• Although if different alleles are fixed in
  different populations can help maintain genetic
  diversity among populations

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SUMMARY

• When one allele is consistently favored it will
  be driven to fixation
• When heterozygote is favored both alleles are
  maintained and at a stable equilibrium (balanced
  polymorphism) even though one of the alleles may
  be lethal in the homozygous state.

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Frequency-dependent selection

• The Elderflower orchid example in book
• Populations allele frequencies remain at or near
  an equilibrium but it is due to the direction of
  selection fluctuating.
• First one allele is favored and then the other.
• The population fluctuates around an equilibrium
  point.

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The favored allele fluctuates because

• Bumblebees visit yellow and purple flowers
  alternately
• The least frequent phenotype is visited more
  often and receives more pollination events.
• In subsequent generations this color becomes more
  and more frequent until it becomes the dominant
  color.
• Once this happens then the same color becomes
  less frequently visited and the other color
  becomes favored.
• Oscillation between the two colors continues and
  the favored allele alternates over time around
  some mean equilibrium value.

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End of Selection Effects
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Effects of Mutation

• Mutation is the source of all new alleles
• Mutation provides the raw material on which
  selection can act

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Hardy Weinberg and Mutation

• Mutation alone is a weak or nonexistent
  evolutionary force
• If all mutations that happened, occurred in
  gametes so that they would be immediately passed
  on to their offspring and .
• the rate of mutation were high, say A?a at a
  rate of 1 in 10,000 per generation.
• then the rates are very slow as shown in figure
  6.23

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Figure 6.23 pg. 211
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Mutation and Selection

• In concert with selection, mutation becomes a
  potent evolutionary force.

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Richard Lenski and colleagues working with E.
coli

• Used a strain of E. coli that cannot exchange DNA
  (conjugation) so the only possible source of
  genetic variation is mutation.
• Showed steady increases in fitness and size over
  10,000 generations in response to a demanding
  environment. (little over 4 years)
• However, increases in fitness occurred in jumps
  when a beneficial mutation occurred and then
  spread rapidly through the population

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Figure 6.25 pg 213
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Mutation Selection Balance

• When mutations are deleterious
• Selection acts to eliminate them
• Deleterious Mutations persist because they are
  created anew over and over again
• When the rate at which deleterious mutations are
  formed exactly equals the rate at which they are
  eliminated by selection the allele is in
  equilibrium. mutation-selection balance

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Intuition tells us that ...

• If the mutation is only mildly deleterious and
  therefore selection against it is weak and
  Mutation rate is high then
• The equilibrium frequency of the mutated allele
  will be relatively in the population.
• If, on the other hand, there is strong selection
  against a mutation (the mutation is highly
  deleterious) and the mutation rate is low then
• Equilibrium ratio of the mutated allele will be
  in the population

high
low
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Example

• Spinal muscular atrophy, second most common
  lethal autosomal recessive disease in humans.
  Selection coefficient is .9 against the disease
  mutations.
• However, among Caucasians 1 in 100 people carry
  the disease causing allele.
• Research shows that the mutation rate for this
  disease is quite high
• Mutation selection balance is proposed
  explanation for persistence of mutant alleles.
• http//www.smafoundation.org

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Cystic Fibrosis

• Cystic fibrosis is the most common lethal
  autosomal recessive disease in Caucasians
• Mutation-selection balance alone cannot account
  for the high frequency of the allele .02
• Appears to also be some heterozygote superiority
  involved
• Heterozygotes are resistant to typhoid fever
  bacteria and have superior fitness during typhoid
  fever epidemic.
• At the current time it is believed that CF is an
  example of heterosis and not mutation-selection
  balance

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Huntingtons Diseasenew research 2007.

• An autosomal dominant allele
• Is actually increasing in the human population.
• Any ideas why?
• May be because it increases the tumor supressor
  activity in cells dramatically lowering the
  incidence of cancer in those with the defective
  allele.
• They survive through the reproductive years and
  leave more offspring than their unaffected
  siblings.