Download - Population Genetics - University of Washingtonevolution.genetics.washington.edu/gs453/2004/section2.pdf · Population Genetics Joe Felsenstein GENOME 453, Winter 2004 Population Genetics

Population Genetics

Joe Felsenstein

GENOME 453, Winter 2004

Population Genetics – p.1/47

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


A Hardy-Weinberg calculation

5 AA 2 Aa 3 aa

0.50 0.20 0.30



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

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



0.6 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



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


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


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


The process of natural selection at one locus

gametes

zygotes

genotypes are lethal in this case



gametes

zygotes




gametes

zygotes

gametes

zygotes




gametes

zygotes

gametes

gametes

zygotes

...



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)



x 1

AA Aa aa1 1 0.7


0.04 0.32 0.64x 1 x 0.7



(newborns)

0.04 0.32 Total: 0.808+ + =0.448

Survivors (these are relative viabilities)



aa

x 1

AA Aa1 1 0.7


0.04 0.32 0.64x 1 x 0.7



(newborns)

0.04 0.32 Total: 0.808

genotype frequencies among the survivors:

+ + =0.448

0.396 0.554


0.0495

(divide by the total)



x 1

AA Aa aa1 1 0.7


0.04 0.32 0.64x 1 x 0.7



(newborns)

0.04 0.32 Total: 0.808


+ + =0.448

0.396 0.554

gene frequency

=

a: 0.7525


0.0495

A:

0.554 + 0.5 x 0.396


0.0495 + 0.5 x 0.396 0.2475

=



x 1

AA Aa aa1 1 0.7


0.04 0.32 0.64x 1 x 0.7



(newborns)

0.04 0.32 Total: 0.808


+ + =0.448

0.396 0.554

gene frequency

=

a: 0.7525


0.0495

A:

0.554 + 0.5 x 0.396


0.0495 + 0.5 x 0.396 0.2475

=

genotype frequencies:

0.0613 0.3725 0.5663

(among newborns)


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)


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


An experimental selection curve


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


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


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__


Genetic drift

Time (generations)

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

Gen

e fr

equ

ency

0

1


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)


A cline (name by Julian Huxley)

no migration

some

more

geographic position

gen

e fr

equ

ency

0

1


A famous common-garden experiment

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


Heavy metal


House sparrows


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


Mutation rates in humans


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


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.


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


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 European

ancestry have the gene. Reduction in fitness maybe 2%.

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

heterozygotes.

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, the

fraction 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.


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.


Download - Population Genetics - University of Washingtonevolution.genetics.washington.edu/gs453/2004/section2.pdf · Population Genetics Joe Felsenstein GENOME 453, Winter 2004 Population Genetics

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