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  • Population Genetics

    Joe Felsenstein

    GENOME 453, Winter 2004

    Population Genetics – p.1/47

  • Godfrey Harold Hardy (1877-1947) Wilhelm Weinberg (1862-1937)

    Population Genetics – p.2/47

  • A Hardy-Weinberg calculation

    5 AA 2 Aa 3 aa

    0.50 0.20 0.30

    Population Genetics – p.3/47

  • A Hardy-Weinberg calculation

    5 AA 2 Aa 3 aa

    0.50 0.20 0.30

    0.50 + (1/2) 0.20 (1/2) 0.20 + 0.30

    Population Genetics – p.4/47

  • A Hardy-Weinberg calculation

    0.6 A

    5 AA 2 Aa 3 aa

    0.50 0.20 0.30

    0.50 + (1/2) 0.20 (1/2) 0.20 + 0.30

    0.6 A 0.4 a

    0.4 a

    Population Genetics – p.5/47

  • A Hardy-Weinberg calculation

    0.6 A

    5 AA 2 Aa 3 aa

    0.50 0.20 0.30

    0.50 + (1/2) 0.20 (1/2) 0.20 + 0.30

    0.6 A 0.4 a

    0.4 a

    0.36 AA 0.24 Aa

    0.24 Aa 0.16 aa

    Population Genetics – p.6/47

  • A Hardy-Weinberg calculation

    0.6 A

    5 AA 2 Aa 3 aa

    0.50 0.20 0.30

    0.50 + (1/2) 0.20 (1/2) 0.20 + 0.30

    0.6 A 0.4 a

    0.4 a

    0.36 AA 0.24 Aa

    0.24 Aa 0.16 aa

    Result:

    0.36 AA

    0.48 Aa0.16 aa

    0.6 A

    0.4 a

    1/2

    1/2

    Population Genetics – p.7/47

  • Calculating the gene frequency (two ways)

    Suppose that we have 200 individuals: 83 AA, 62 Aa, 55 aa

    Method 1. Calculate what fraction of gametes bear A:

    AA

    Aa

    aa

    Genotype Number

    83

    62

    55

    Genotype frequency

    0.415

    0.31

    0.275

    Fraction of gametes

    0.57

    0.43

    1/2

    1/2

    A

    a

    all

    all

    Population Genetics – p.8/47

  • Calculating the gene frequency (two ways)

    AA

    Aa

    aa

    Genotype Number

    83

    62

    55

    0.57

    0.43

    A

    a

    A’s a’s

    0

    1100

    166

    62 62

    228 172+ = 400

    228

    400=

    400

    172=

    Method 2. Calculate what fraction of genes in the parents are A:

    Suppose that we have 200 individuals: 83 AA, 62 Aa, 55 aa

    Population Genetics – p.9/47

  • The process of natural selection at one locus

    gametes

    zygotes

    genotypes are lethal in this case

    Population Genetics – p.10/47

  • The process of natural selection at one locus

    gametes

    zygotes

    genotypes are lethal in this case

    Population Genetics – p.11/47

  • The process of natural selection at one locus

    gametes

    zygotes

    gametes

    zygotes

    genotypes are lethal in this case

    Population Genetics – p.12/47

  • The process of natural selection at one locus

    gametes

    zygotes

    gametes

    zygotes

    genotypes are lethal in this case

    Population Genetics – p.13/47

  • The process of natural selection at one locus

    gametes

    zygotes

    gametes

    gametes

    zygotes

    ...

    genotypes are lethal in this case

    Population Genetics – p.14/47

  • The process of natural selection at one locus

    gametes

    zygotes

    gametes

    gametes

    zygotes

    ...

    genotypes are lethal in this case

    Population Genetics – p.15/47

  • The process of natural selection at one locus

    gametes

    zygotes

    gametes

    gametes

    zygotes

    ...

    genotypes are lethal in this case

    Population Genetics – p.16/47

  • The process of natural selection at one locus

    gametes

    zygotes

    gametes

    gametes

    zygotes

    ...

    genotypes are lethal in this case

    Population Genetics – p.17/47

  • The process of natural selection at one locus

    gametes

    zygotes

    gametes

    gametes

    zygotes

    ...

    genotypes are lethal in this case

    Population Genetics – p.18/47

  • The process of natural selection at one locus

    gametes

    zygotes

    gametes

    gametes

    zygotes

    ...

    genotypes are lethal in this case

    Population Genetics – p.19/47

  • The process of natural selection at one locus

    gametes

    zygotes

    gametes

    gametes

    zygotes

    ...

    genotypes are lethal in this case

    Population Genetics – p.20/47

  • The process of natural selection at one locus

    gametes

    zygotes

    gametes

    gametes

    zygotes

    ...

    genotypes are lethal in this case

    Population Genetics – p.21/47

  • A numerical example of natural selection

    AA Aa aa1 1 0.7

    Initial gene frequency of A = 0.2Initial genotype frequencies (from Hardy−Weinberg)

    0.04 0.32 0.64

    Genotypes:relativefitnesses:

    (assume these are viabilities)

    (newborns)

    Population Genetics – p.22/47

  • A numerical example of natural selection

    x 1

    AA Aa aa1 1 0.7

    Initial gene frequency of A = 0.2Initial genotype frequencies (from Hardy−Weinberg)

    0.04 0.32 0.64x 1 x 0.7

    Genotypes:relativefitnesses:

    (assume these are viabilities)

    (newborns)

    0.04 0.32 Total: 0.808+ + =0.448

    Survivors (these are relative viabilities)

    Population Genetics – p.23/47

  • A numerical example of natural selection

    aa

    x 1

    AA Aa1 1 0.7

    Initial gene frequency of A = 0.2Initial genotype frequencies (from Hardy−Weinberg)

    0.04 0.32 0.64x 1 x 0.7

    Genotypes:relativefitnesses:

    (assume these are viabilities)

    (newborns)

    0.04 0.32 Total: 0.808

    genotype frequencies among the survivors:

    + + =0.448

    0.396 0.554

    Survivors (these are relative viabilities)

    0.0495

    (divide by the total)

    Population Genetics – p.24/47

  • A numerical example of natural selection

    x 1

    AA Aa aa1 1 0.7

    Initial gene frequency of A = 0.2Initial genotype frequencies (from Hardy−Weinberg)

    0.04 0.32 0.64x 1 x 0.7

    Genotypes:relativefitnesses:

    (assume these are viabilities)

    (newborns)

    0.04 0.32 Total: 0.808

    genotype frequencies among the survivors:

    + + =0.448

    0.396 0.554

    gene frequency

    =

    a: 0.7525

    Survivors (these are relative viabilities)

    0.0495

    A:

    0.554 + 0.5 x 0.396

    (divide by the total)

    0.0495 + 0.5 x 0.396 0.2475

    =

    Population Genetics – p.25/47

  • A numerical example of natural selection

    x 1

    AA Aa aa1 1 0.7

    Initial gene frequency of A = 0.2Initial genotype frequencies (from Hardy−Weinberg)

    0.04 0.32 0.64x 1 x 0.7

    Genotypes:relativefitnesses:

    (assume these are viabilities)

    (newborns)

    0.04 0.32 Total: 0.808

    genotype frequencies among the survivors:

    + + =0.448

    0.396 0.554

    gene frequency

    =

    a: 0.7525

    Survivors (these are relative viabilities)

    0.0495

    A:

    0.554 + 0.5 x 0.396

    (divide by the total)

    0.0495 + 0.5 x 0.396 0.2475

    =

    genotype frequencies:

    0.0613 0.3725 0.5663

    (among newborns)

    Population Genetics – p.26/47

  • The algebra of natural selection

    mean fitness of A

    mean fitnessof everybody

    AA

    p2

    wAA

    p2

    wAA

    Aa

    2pq

    wAa

    2pq wAa

    aa

    q2

    waa

    q2

    waa

    Genotype:

    After selection:

    Frequency:

    Relative fitnesses:

    Note that these don’t add up to 1

    New gene frequency is then

    = p

    2w

    AA2pq w

    Aa

    p2

    wAA

    2pq wAa

    q2

    waa

    + (1/2)

    + +

    p’

    = p w

    AAq w

    Aa

    p2

    wAA

    2pq wAa

    q2

    waa

    +

    + +

    p’p ( )

    = pw

    A

    w

    (adding up A bearers and dividing by everybody)

    Population Genetics – p.27/47

  • Is weak selection effective?

    Suppose (relative) fitnesses are:

    AA Aa aa

    (1+s)2 1+s 1

    x (1+s) x (1+s)

    So in this example each change ofa to A multiplies the fitnessby (1+s), so that it increases itby a fraction s.

    0.01 − 0.1 0.1 − 0.5 0.5 − 0.9 0.9 − 0.99s

    The time for gene frequency change, in generations, turns out to be:

    change of gene frequencies

    1

    0.1

    0.01

    0.001

    3.46 3.17 3.17 3.46

    25.16 23.05 23.05 25.16

    240.99 220.82 220.82 240.99

    2399.09 2198.02 2198.02 2399.09

    Ag

    ene

    freq

    uen

    cy o

    f

    generations

    0

    1

    0.5

    Population Genetics – p.28/47

  • An experimental selection curve

    Population Genetics – p.29/47

  • Rare alleles occur mostly in heterozygotes

    Genotype frequencies:

    0.81 AA : 0.18 Aa : 0.01 aa

    Note that of the 20 copies of a,

    18 of them, or 18 / 20 = 0.9 of them are in Aa genotypes

    This shows a population in Hardy−Weinberg equilibrium

    at gene frequencies of 0.9 A : 0.1 a

    Population Genetics – p.30/47

  • Overdominance and polymorphism

    AA Aa aa1 − s 1 1 − t

    when A is rare, most A’s are in Aa, and most a’s are in aa

    The average fitness of A−bearing genotypes is then nearly 1

    The average fitness of a−bearing genotypes is then nearly 1−t

    So A will increase in frequency when rare

    when a is rare, most a’s are in Aa, and most A’s are in AA

    The average fitness of a−bearing genotypes is then nearly 1

    The average fitness of A−bearing genotypes is then nearly 1−s

    So a will increase in frequency when rare

    0 1gene frequency of A

    Population Genetics – p.31/47

  • Underdominance and unstable equilibrium

    AA Aa aa1

    when A is rare, most A’s are in Aa, and most a’s are in aa

    The average fitness of A−bearing genotypes is then nearly 1

    when a is rare, most a’s are in Aa, and most A’s are in AA

    The average fitness of a−bearing genotypes is then nearly 1

    1gene frequency of A

    1+s 1+t

    The average fitness of a−bearing genotypes is then nearly 1+t

    So A will decrease in frequency when rare

    The average fitness of A−bearing genotypes is then nearly 1+s

    So a will decrease in frequency when rare

    0Population Genetics – p.32/47

  • Fitness surfaces (adaptive landscapes)

    Overdominance

    0 1p0 1p

    stable equilibrium

    (gene frequency changes)

    Underdominance

    w__

    0 1p

    (gene frequency changes)

    unstable equilibrium

    Is all for the best in this best of all possible worlds?

    Can you explain the underdominance result in terms of rare alleles beingmostly in heterozygotes?

    w__

    Population Genetics – p.33/47

  • Genetic drift

    Time (generations)

    0 1 2 3 4 5 6 7 8 9 10 11

    Gen

    e fr

    equ

    ency

    0

    1

    Population Genetics – p.34/47

  • Distribution of gene frequencies with drift

    0 1

    0 1

    0 1

    0 1

    0 1

    time

    Note that although the individual populationswander their average hardly moves (not at allwhen we have infinitely many populations)

    Population Genetics – p.35/47

  • A cline (name by Julian Huxley)

    no migration

    some

    more

    geographic position

    gen

    e fr

    equ

    ency

    0

    1

    Population Genetics – p.36/47

  • A famous common-garden experiment

    Clausen, Keck and Hiesey’s (1949) common-garden experiment inAchillea lanulosa

    Population Genetics – p.37/47

  • Heavy metal

    Population Genetics – p.38/47

  • House sparrows

    Population Genetics – p.39/47

  • House sparrows

    Population Genetics – p.40/47

  • Mutation Rates

    Coat color mutants in mice. From

    Schlager G. and M. M. Dickie. 1967. Spontaneous mutations and

    mutation rates in the house mouse. Genetics 57: 319-330

    Locus Gametes tested No. of Mutations Rate

    Nonagouti 67,395 3 4.4 × 10−6

    Brown 919,619 3 3.3 × 10−6

    Albino 150,391 5 33.2 × 10−6

    Dilute 839,447 10 11.9 × 10−6

    Leaden 243,444 4 16.4 × 10−6

    ——- — ————-Total 2,220,376 25 11.2 × 10−6

    Population Genetics – p.41/47

  • Mutation rates in humans

    Population Genetics – p.42/47

  • Forward vs. back mutations

    Why mutants inactivating a functional gene will be more

    frequent than back mutations

    The gene

    12 places can mutate to nonfunctionality

    only one place can mutate back to function

    function can sometimes be restored by a "second site" mutation, too

    Population Genetics – p.43/47

  • A sequence space

    For sequences of length 1000, there are 3 X 1000 = 3000 "neighbors" one step away in sequence space

    Hard to draw such a space

    How do we ever evolve? Woiuldn’t it be impossible to find one of the tiny fraction of

    possible sequences that would be even marginally functional?

    The answer seems to be that the sequences are clustered

    An example of such clustering is the English language, as illustrated by a popular word game:

    There are also only a tiny fraction of all 456,976 four−letter words that are English words

    But they are clustered, so that it is possible to "evolve" from one to another through intermediates

    W O R D

    W O R E

    G O R E

    G O N E

    G E N E

    But there are 4 1000 602 sequences, which is about 10 in all !

    But the word BCGHcannot be made intoan English word

    No two of them are more than 1000 steps apart.

    Population Genetics – p.44/47

  • Mutation as an evolutionary force

    10−6

    and mutation rate back is the same,

    0

    1

    0 1 million generations

    0.5

    Mutation is critical in introducing new alleles but is very slow

    in changing their frequencies

    If we have two alleles A and a, and mutation rate from A to a is

    Population Genetics – p.45/47

  • Estimation of a human mutation rate

    By an equilibrium calculation. Huntington’s disease. Dominant. Does

    not express itself until after age 40. 1/100, 000 of people of Europeanancestry have the gene. Reduction in fitness maybe 2%.

    If allele frequency is q, then 2q(1 − q) of everyone areheterozygotes.

    0.02 of these die. Each has half its copies the Huntington’s allele.

    So as the frequency of people with the gene is ' 1/100, 000, thefraction of all copies that are mutations that are eliminated is

    0.00001 × 1/2 × 0.02 ' 10−7

    If we are at equilibrium between mutation and selection, this isalso the fraction of copies that have a new mutation.

    Similar calculations can be done with recessive alleles.

    Population Genetics – p.46/47

  • How it was done

    This projection produced

    using the prosper style in LaTeX,

    using Latex to make a .dvi file,

    using dvips to turn this into a Postscript file,

    using ps2pdf to mill a PDF file, and

    displaying the slides in Adobe Acrobat Reader.

    Result: nice slides using freeware.

    Population Genetics – p.47/47

    A Hardy-Weinberg calculationA Hardy-Weinberg calculationA Hardy-Weinberg calculationA Hardy-Weinberg calculationA Hardy-Weinberg calculationCalculating the gene frequency (two ways)Calculating the gene frequency (two ways)The process of natural selection at one locusThe process of natural selection at one locusThe process of natural selection at one locusThe process of natural selection at one locusThe process of natural selection at one locusThe process of natural selection at one locusThe process of natural selection at one locusThe process of natural selection at one locusThe process of natural selection at one locusThe process of natural selection at one locusThe process of natural selection at one locusThe process of natural selection at one locusA numerical example of natural selectionA numerical example of natural selectionA numerical example of natural selectionA numerical example of natural selectionA numerical example of natural selectionThe algebra of natural selectionIs weak selection effective?An experimental selection curveRare alleles occur mostly in heterozygotesOverdominance and polymorphismUnderdominance and unstable equilibriumFitness surfaces (adaptive landscapes)Genetic driftDistribution of gene frequencies with driftA cline (name by Julian Huxley)A famous common-garden experimentHeavy metalHouse sparrowsHouse sparrowsMutation RatesMutation rates in humansForward vs. back mutationsA sequence spaceMutation as an evolutionary forceEstimation of a human mutation rateHow it was done