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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 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.
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