Population Genetics: Selection and mutation as mechanisms of evolution

1
Population Genetics Selection and mutation as
mechanisms of evolution

• Population genetics study of Mendelian genetics
  at the level of the whole population.

2
Hardy-Weinberg Equilibrium

• To understand the conditions under which
  evolution can occur, it is necessary to
  understand the population genetic conditions
  under which it will not occur.

3
Hardy-Weinberg Equilibrium

• The Hardy-Weinberg Equilibrium Principle allows
  us do this. It enables us to predict allele and
  genotype frequencies from one generation to the
  next in the absence of evolution.

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Hardy-Weinberg Equilibrium

• Assume a gene with two alleles A and a with known
  frequencies (e.g. A 0.6, a 0.4.)
• There are only two alleles in the population so
  their frequencies must add up to 1.

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Hardy-Weinberg Equilibrium

• Using allele frequencies we can predict the
  expected frequencies of genotypes in the next
  generation.
• With two alleles only three genotypes are
  possible AA, Aa and aa

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Hardy-Weinberg Equilibrium

• Assume alleles A and a are present in eggs and
  sperm in proportion to their frequency in
  population (i.e. 0.6 and 0.4)
• Also assume that sperm and eggs meet at random
  (one big gene pool).

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Hardy-Weinberg Equilibrium

• Then we can calculate expected genotype
  frequencies.
• AA To produce an AA individual, egg and sperm
  must each contain an A allele.
• This probability is 0.6 x 0.6 0.36 (probability
  sperm contains A times probability egg contains
  A).

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Hardy-Weinberg Equilibrium

• Similarly, we can calculate frequency of aa.
• 0.4 x 04 0.16.

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Hardy-Weinberg Equilibrium

• Probability of Aa is given by probability sperm
  contains A (0.6) times probability egg contains a
  (0.4). 0.6 x 04 0.24.

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Hardy-Weinberg Equilibrium

• But, theres a second way to produce an Aa
  individual (egg contains A and sperm contains a).
  Same probability as before 0.6 x 0.4 0.24.
• Hence the overall probability of Aa 0.24 0.24
  0.48.

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Hardy-Weinberg Equilibrium

• Genotypes in next generation
• AA 0.36
• Aa 0.48
• Aa 0.16
• These frequencies add up to one.

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Hardy-Weinberg Equilibrium

• General formula for Hardy-Weinberg.
• Let p frequency of allele A and q frequency of
  allele a.
• p2 2pq q2 1.

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Hardy Weinberg Equilibrium with more than 2
alleles

• If three alleles with frequencies P1, P2 and P3
  such that P1 P2 P3 1
• Then genotype frequencies given by
• P12 P22 P32 2P1P2 2P1 P3
• 2P2P3

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Conclusions from Hardy-Weinberg Equilibrium

• Allele frequencies in a population will not
  change from one generation to the next just as a
  result of assortment of alleles and zygote
  formation.
• Assortment of alleles simply means what occurs
  during meiosis when only one copy of each pair of
  alleles enters any given gamete (remember each
  gamete only contains half the DNA of a body
  cell).

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Conclusions from Hardy-Weinberg Equilibrium

• If the allele frequencies in a gene pool with two
  alleles are given by p and q, the genotype
  frequencies will be given by p2, 2pq, and q2.

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Working with the H-W equation

• You need to be able to work with the
  Hardy-Weinberg equation.
• For example, if 9 of 100 individuals in a
  population suffer from a homozygous recessive
  disorder can you calculate the frequency of the
  disease-causing allele? Can you calculate how
  many heterozygotes are in the population?

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Working with the H-W equation

• p2 2pq q2 1. The terms in the equation
  represent the frequencies of individual
  genotypes. A genotype is possessed by an
  individual organism so there are two alleles
  present in each case.
• P and q are allele frequencies. Allele
  frequencies are estimates of how common alleles
  are in the whole population.
• It is vital that you understand the difference
  between allele and genotye frequencies.

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Working with the H-W equation

• 9 of 100 (frequency 0.09) of individuals are
  homozygous for the recessive allele. What term in
  the H-W equation is that equal to?

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Working with the H-W equation

• Its q2.
• If q2 0.09, whats q? Get square root of q2,
  which is 0.3, which is the frequency of the
  allele a.
• If q0.3 then p0.7. Now plug p and q into
  equation to calculate frequencies of other
  genotypes.

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Working with the H-W equation

• p2 (0.7)(0.7) 0.49 -- frequency of AA
• 2pq 2 (0.3)(0.7) 0.42 frequency of Aa.
• To calculate the actual number of heterozygotes
  simply multiply 0.42 by the population size
  (0.42)(100) 42.

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Working with the H-W equation 3 alleles

• There are three alleles in a population A1, A2
  and A3 whose frequencies respectively are 0.2,
  0.2 and 0.6 and there are 100 individuals in the
  population.
• How many A1A2 heterozygotes will there be in the
  population?

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Working with the H-W equation 3 alleles

• Just use the formulae P1 P2 P3 1 and P12
  P22 P32 2P1P2 2P1 P3
• 2P2P3 1
• Then substitute in the appropriate values for the
  appropriate term
• 2P1P2 2(0.2)(0.2) 0.08 or 8 people out of
  100.

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Other examples of working with HW equilibrium is
a population in HW equilibrium?

• In a population there are 100 birds with the
  following genotypes
• 44 AA
• 32 Aa
• 24 aa
• How would you demonstrate that this population is
  not in Hardy Weinberg equilibrium

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Three steps

• Step 1 calculate the allele frequencies.
• Step 2 Calculate expected numbers of each
  geneotype (i.e. figure out how many homozygotes
  and heterozygotes you would expect.)
• Step 3 compare your expected and observed data.
  

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Step 1 allele frequencies

• Step 1. How many A alleles are there in total?
• 44 AA individuals 88 A alleles (because each
  individual has two copies of the A allele)
• 32 Aa alleles 32 A alleles
• Total A alleles is 8832 120.

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Step 1 allele frequencies

• Total number of a alleles is similarly
  calculated as 224 32 80
• What are allele frequencies?
• Total number of alleles in population is 120 80
  200 (or you could calculate it by multiplying
  the number of individuals in the population by
  two 1002 200)

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Step 1 allele frequencies

• Allele frequencies are
• A 120/200 0.6. Let p 0.6
• a 80/200 0.4. Let q 0.4

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Step 2 Calculate expected number of each genotype

• Use the Hardy_Weinberg equation
• p2 2pq q2 1 to calculate what expected
  genotypes we should have given these observed
  frequencies of A and a
• Expected frequency of AA p2 0.6 0.6 0.36
• Expected frequency of aa q2 0.40 .4 0.16
  
• Expected frequency of Aa 2pq 2.6.4 0.48

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Step 2 Calculate expected number of each genotype

• Convert genotype frequencies to actual numbers by
  multiplying by population size of 100
• AA 0.36100 36
• aa 0.16100 16
• Aa 0.48100 48

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Step 3 Compare Observed and Expected values

• Observed population is
• 44 AA 32 Aa 24 aa
• Expected population is
• 36AA 48Aa 16aa
• These numbers are not the same so the population
  is not in Hardy-Weinberg equilibrium. An
  assumption of the Hardy Weinberg equilibrium is
  being violated. What are those assumptions?

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Assumptions of Hardy-Weinberg

• 1. No selection.
• When individuals with certain genotypes survive
  better than others, allele frequencies may change
  from one generation to the next.

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Assumptions of Hardy-Weinberg

• 2. No mutation
• If new alleles are produced by mutation or
  alleles mutate at different rates, allele
  frequencies may change from one generation to the
  next.

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Assumptions of Hardy-Weinberg

• 3. No migration
• Movement of individuals in or out of a population
  will alter allele and genotype frequencies.

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Assumptions of Hardy-Weinberg

• 4. No chance events.
• Luck plays no role. Eggs and sperm collide at
  same frequencies as the actual frequencies of p
  and q.
• When assumption is violated, and by chance some
  individuals contribute more alleles than others
  to next generation, allele frequencies may
  change. This mechanism of allele frequency
  change is called Genetic Drift.

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Assumptions of Hardy-Weinberg

• 5. Individuals select mates at random.
• If this assumption is violated allele frequencies
  will not change, but genotype frequencies may.

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Hardy-Weinberg Equilibrium

• Hardy Weinberg equilibrium principle identifies
  the forces that can cause evolution.
• If a population is not in H-W equilibrium then
  one or more of the five assumptions is being
  violated.

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5.10
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Can selection change allele frequencies?

• Two alleles B1 and B2
• Frequency of B1 0.6 and frequency of B2 0.4.
• Random mating gives genotype frequencies 0.36
  B1B1 0.48B1B2 0.16B2B2
  

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Can selection change allele frequencies?

• Assume 100 individuals
• 36 B1B1 48 B1B2 16 B2B2
• Incorporate selection. Assume all B1B1 survive,
  75 of B1B2 survive and 50 of B2B2 survive.

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Can selection change allele frequencies?

• Population now 80 individuals 36 B1B1 36 B1B2
  8 B2B2
• Allele frequencies now
• B1 72 36/160 0.675
• B2 3616/160 0.325
• Selection resulted in allele frequency change.

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FIG 5.11
42
Can selection change allele frequencies?

• Selection in previous example very strong.
• What patterns expected with weaker selection.
• Initial frequencies B1 0.01, B2 0.99.

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Fig 5.12
Note line colors yellow and red have been flipped
in the table.
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Can selection change allele frequencies?

• Rate of change of B1is rapid when selection
  pressure is strong, but much slower, although
  still steady, under weak selection.

45
Empirical examples of allele frequency change
under selection

• Clavener and Cleggs work on Drosophila.
• Two alleles for ADH (alcohol dehydrogenase breaks
  down ethanol) ADHF and ADHS

46
Empirical examples of allele frequency change
under selection

• Two Drosophila populations maintained one fed
  food spiked with ethanol, control fed unspiked
  food.
• Populations maintained for multiple generations.

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Empirical examples of allele frequency change
under selection

• Experimental population showed consistent
  long-term increase in frequency of ADHF
• Flies with ADHF allele have higher fitness when
  ethanol is present.
• ADHF enzyme breaks down ethanol twice as fast as
  ADHS enzyme.

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Fig 5.13
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Empirical examples of allele frequency change
under selection Jaeken syndrome

• Jaeken syndrome patients severely disabled with
  skeletal deformities and inadequate liver
  function.

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Jaeken syndrome

• Autosomal recessive condition caused by
  loss-of-function mutation of gene PMM2 codes for
  enzyme phosphomannomutase.
• Patients unable to join carbohydrates and
  proteins to make glycoproteins at a high enough
  rate.
• Glycoproteins involved in movement of substances
  across cell membranes.

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Jaeken syndrome

• Many different loss-of-function mutations can
  cause Jaeken Syndrome.
• Team of researchers led by Jaak Jaeken
  investigated whether different mutations differed
  in their severity. Used Hardy-Weinberg
  equilibrium to do so.

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Jaeken syndrome

• People with Jaeken syndrome are homozygous for
  the disease, but may be either homozygous or
  heterozygous for a given disease allele.
• Different disease alleles should be in
  Hardy-Weinberg equilibrium.

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Jaeken syndrome

• Researchers studied 54 patients and identified
  most common mutation as R141H.
• Dividing population into R141H and other alleles
  Other. Allele frequencies are
• Other 0.6 and R141H 0.4.

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Jaeken syndrome

• If disease alleles are in H-W equilibrium then we
  predict genotype frequencies of
• Other/other 0.36
• Other/R141H 0.48
• R141H/R141H 0.16

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Jaeken syndrome

• Observed frequencies are
• Other/Other 0.2
• Other/R141H 0.8
• R141H/R141H 0
• Clearly population not in H-W equilibrium.

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Jaeken syndrome

• Researchers concluded that R141H is an especially
  severe mutation and homozygotes die before or
  just after birth.
• Thus, there is selection so H-W assumption is
  violated.

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Using H-W to predict potential spread of CCR5-?32
allele

• Will AIDS epidemic cause CCR5-?32 (delta32)
  allele to spread? Offers protection against HIV
  infection.
• In principle it could, but models suggest it
  probably will not in any real population.

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Spread of CCR5-?32 allele

• Scenario 1. Initial allele frequency 20. 25
  of heterozygotes and those homozygous for normal
  allele die of AIDS. Homozygous ?32 individuals
  do not die of AIDS.
• Over 40 generations allele increases to almost
  100.

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FIG 5.15a
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Spread of CCR5-?32 allele

• In human population with high HIV infection rate
  and high frequencies of ?32 allele, ?32 could
  spread rapidly.

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Spread of CCR5-?32 allele

• Scenario 2. In areas with low HIV infection
  rates (under 1), but high levels of ?32 (as
  found in Europe), selection too weak to raise
  delta 32 frequency much.

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FIG 5.15b
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Spread of CCR5-?32 allele

• Scenario 3. In sub-Saharan Africa HIV infection
  rate about 25. However, ?32 allele almost
  absent
• Under these conditions ?32 frequency will hardly
  change because most copies of allele are in
  heterozygotes, which are not protected from HIV.

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FIG 5.15c
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Testing predictions of population genetics theory

• Theory predicts that if an individual carrying an
  allele has higher than average fitness then the
  frequency of that allele will increase from one
  generation to the next.
• Obviously, the converse should be true and a
  deleterious allele should decrease in frequency
  if its bearers have lower fitness.

66
Testing predictions of population genetics theory

• Mathematical treatment of effect of selection on
  gene frequencies is given in Box 6.3 (page 186)
  of your text.
• Main point is that if the average fitness of an
  allele A when paired at random with other alleles
  in the population is higher than the average
  fitness of the population, then it will increase
  in frequency.

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Tests of theory

• Dawson (1970). Flour beetles. Two alleles at
  locus and l.
• / and /l phenotypically normal.
• l/l lethal.

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Dawsons flour beetles

• Dawson founded two populations with heterozygotes
  (frequency of and l alleles thus 0.5).
• Expected allele to increase in frequency and l
  allele to decline over time.

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Fig 5.16a
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Dawsons flour beetles

• Predicted and observed allele frequencies matched
  very closely.
• l allele declined rapidly at first, but rate of
  decline slowed.

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Dawsons flour beetles

• Dawsons results show that when a recessive
  allele is common, evolution by natural selection
  is rapid, but slows as recessive allele becomes
  rarer.
• Hardy-Weinberg explains why.

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Dawsons flour beetles

• When recessive allele (a) common e.g. freq a0.95
  genotype frequencies are
• AA (0.05)2 Aa (2 (0.05)(0.95) aa (0.95)2
• 0.0025AA 0.095Aa 0.9025aa
• With more than 90 of phenotypes being recessive,
  if aa is selected against we expect rapid
  population change.

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Dawsons flour beetles

• When recessive allele (a) rare e.g. freq a0.05
  genotype frequencies are
• AA (0.95)2 Aa 2(0.95)(0.05) aa (0.05)2
• 0.9025AA 0.095Aa 0.0025aa
• Fewer than 0.25 of phenotypes are aa recessive.
  Most a alleles (gt97) are hidden from selection
  as heterozygotes.
• Expect only slow change in frequency of a.

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Maintaining multiple alleles in gene pool

• Dawsons beetle work shows that deleterious rare
  alleles may be very hard to eliminate from a gene
  pool because they remain hidden from selection as
  heterozygotes.
• (this only applies if the allele is not dominant.)

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Maintaining multiple alleles in gene pool

• There are several different ways in which
  multiple alleles may be maintained in populations
  even, as we saw, if one allele is deleterious.
• 1. Deleterious recessive allele can hide in
  heterozygote
• 2. There may be heterozygote advantage
• 3. Frequency-dependent selection

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Maintaining multiple alleles in gene pool-2.
Heterozygote advantage

• Another way in which multiple alleles may be
  maintained in a population is through
  heterozygote advantage.
• Classic example is sickle cell allele.

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Sickle Cell Anemia

• Sickle cell anemia is a condition common among
  West Africans (and African Americans of West
  African ancestry).
• In sickle cell anemia red blood cells are
• sickle shaped.
• Usually fatal by about age 10.

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About 1 of West Africans have sickle cell
anemia. A single mutation that causes a
valine amino acid to replace a glutamine in an
alpha chain of the hemoglobin molecule. Mutation
causes molecules to stick together.
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Why isnt mutant sickle cell gene eliminated by
natural selection?
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Only individuals homozygous for sickle cell gene
get sickle cell anemia. Individuals with one
copy of sickle cell gene (heterozygotes) get
sickle cell trait (mild form of
disease). Individuals with sickle cell allele
(one or two copies) dont get malaria.
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Heterozygotes have higher survival than either
homozygote. Heterozygote advantage. Sickle cell
homozygotes die of sickle cell anemia. Normal
homozygotes more likely to die of
malaria. Stabilizing selection for sickle cell
allele.
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Maintaining multiple alleles in gene pool- 3.
Frequency-dependent selection

• Another way in which multiple alleles are
  maintained is frequency-dependent selection.
• Frequency-dependent selection occurs when rare
  alleles have a selective advantage because they
  are rare.

87
Frequency-dependent selection

• Color polymorphism in Elderflower Orchid
• Two flower colors yellow and purple. Offer no
  food reward to bees. Bees alternate visits to
  colors.
• How are two colors maintained in the population?

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Frequency-dependent selection

• Gigord et al. hypothesis Bees tend to visit
  equal numbers of each flower color. As a result
  the rarer color will have an advantage (because
  each individual rarer color flower will get more
  visits from pollinators).

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Frequency-dependent selection

• Experiment provided five arrays of potted
  orchids with different frequencies of yellow
  orchids in each.
• Monitored orchids for fruit set and removal of
  pollinaria (pollen bearing structures)

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Frequency-dependent selection

• As predicted, reproductive success of yellow
  varied with frequency.

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5.21 a
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Mutation as an evolutionary force

• It is obvious that selection is a very powerful
  evolutionary force but how strong is mutation
  alone as an evolutionary force?
• To check Two alleles A and a.
• Frequency of A 0.9, a 0.1.

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Mutation as an evolutionary force

• Assume A mutates to a at rate of 1 copy per
  10,000 per generation (high rate, but within
  observed range) and all mutations occur in
  gametes.
• How much does this change gene pool in next
  generation?

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Mutation as an evolutionary force

• Hardy Weinberg genotypes in current generation
• 0.81 AA, 0.18 Aa, 0.01 aa
• With no mutation allele frequency in gene pool
  0.9 A, 0.1 a

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Mutation as an evolutionary force

• But mutation reduces frequency of A and increases
  frequency of a
• A a
• 0.9 - (0.0001)(0.9) 0.1 (0.0001)(0.9)
• 0.89991A 0.10009a

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5.23
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Mutation as an evolutionary force

• Not a big change.
• After 1000 generations frequency of A 0.81.

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5.24
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Mutation as an evolutionary force

• Mutation alone clearly not a powerful
  evolutionary force.
• But mutation AND selection together make a very
  powerful evolutionary force.
• Remember mutation provides the raw material for
  selection to work with.

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Lenskis E. coli work

• Lenski et al. studied mutation and selection
  together in an E. coli strain that did not
  exchange DNA (hence mutation only source of new
  variation).
• Bacteria grown in challenging environment (low
  salts and low glucose medium) so selection would
  be strong.

101
Lenskis E. coli work

• 12 replicate populations tracked over about
  10,000 generations.
• Fitness and cell size of populations increased
  over time.
• Pattern of change interesting steplike.
• Why is it steplike?

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5.25
103
Lenskis E. coli work

• Step-like pattern results when a new mutation
  occurs and sweeps through the population as
  mutant bacteria out-reproduce competitors.
• Remember, without mutation evolution would
  eventually cease. Mutation is the ultimate source
  of genetic variation.

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Mutation-selection balance

• Most mutations are deleterious and natural
  selection acts to remove them from population.
• Deleterious alleles persist, however, because
  mutation continually produces them.

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Mutation-selection balance

• When rate at which deleterious alleles are being
  eliminated is equal to their rate of production
  by mutation we have mutation-selection balance.

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Mutation-selection balance

• Equilibrium frequency of deleterious allele q
  square root of µ/s where µ is mutation rate and s
  is the selection coefficient (measure of strength
  of selection against allele ranges from 0 to 1).
  See Box 6.10 page 215 for derivation of equation.

107
Mutation-selection balance

• Equation makes intuitive sense.
• If s is small (mutation only mildly deleterious)
  and µ (mutation rate) is high than q (allele
  frequency) will also be relatively high.
• If s is large and µ is low, than q will be low
  too.

108
Mutation-selection balance

• Spinal muscular atrophy is a generally lethal
  condition caused by a mutation on chromosome 5.
• Selection coefficient estimated at 0.9.
  Deleterious allele frequency about 0.01 in
  Caucasians.
• Inserting above numbers into equation and solving
  for µ get estimated mutation rate of 0.9 X 10-4

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Mutation-selection balance

• Observed mutation rate is about 1.1 X10-4, very
  close agreement in estimates.
• High frequency of allele is accounted for by
  observed mutation rate.

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Is frequency of Cystic fibrosis maintained by
mutation selection balance?

• Cystic fibrosis is caused by a loss of function
  mutation at locus on chromosome 7 that codes for
  CFTR protein (cell surface protein in lungs and
  intestines).
• Major function of protein is to destroy
  Pseudomonas aeruginosa bacteria. Bacterium causes
  severe lung infections in CF patients.

111
Cystic fibrosis

• Very strong selection against CF alleles, but CF
  frequency about 0.02 in Europeans.
• Can mutation rate account for high frequency?

112
Cystic fibrosis

• Assume selection coefficient (s) of 1 and q
  0.02.
• Estimate mutation rate µ is 4.0 X 10-4
• But actual mutation rate is only 6.7 X 10-7
• Is there an alternative explanation?

113
Cystic fibrosis

• May be heterozygote advantage.
• Pier et al. (1998) hypothesized CF heterozygotes
  may be resistant to typhoid fever.
• Typhoid fever caused by Salmonella typhi
  bacteria. Bacteria infiltrate gut by crossing
  epithelial cells.

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Cystic fibrosis

• Hypothesized that S. typhi bacteria may use CFTR
  protein to enter cells.
• If so, CF-heterozygotes should be less vulnerable
  to S. typhi because their gut epithilial cells
  have fewer CFTR proteins on cell surface.

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Cystic fibrosis

• Experimental test.
• Produced mouse cells with three different CFTR
  genotypes
• CFTR homozygote (wild type)
• CFTR/?F508 heterozygote (?F508 most common CF
  mutant allele)
• ?F508/?F508 homozygote

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Cystic fibrosis

• Exposed cells to S. typhi bacteria.
• Measured number of bacteria that entered cells.
• Clear results

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Fig 5.27a
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Cystic fibrosis

• ?F508/?F508 homozygote almost totally resistant
  to S. typhi.
• Wild type homozygote highly vulnerable
• Heterozygote contained 86 fewer bacteria than
  wild type.

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Cystic fibrosis

• Further support for idea ?F508 provides
  resistance to typhoid provided by positive
  relationship between ?F508 allele frequency in
  generation after typhoid outbreak and severity of
  the outbreak.

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Fig 5.27b
Data from 11 European countries