Download 6A - Selection - A Quantitative Look

A Brief Look at Population
Genetics and the Quantitative
Study of Natural Selection
Population Genetics
• Concerned with genotype and gene frequencies
• Vast numbers of loci; human genome may contain 100,000 loci
• Easiest to consider models of evolution at one locus
Consider a population of eight individuals
• Aa, AA, aa, aa, AA, Aa, AA, Aa
• To find the genotype frequencies, simply
count the number of individuals with each
genotype
• So, frequency of AA = 3/8 = 0.375, Aa =
3/8 = 0.375, and aa = 2/8 = 0.25
What about gene frequencies?
A
a
A
A
a
a
a
a
A
A
A
a
A
A
A
a
Each genotype contains two
genes, so there are a total of 16
genes per locus in a population
of eight individuals.
The frequency of A is 9/16 or 0.5625,
while the frequency of a is 7/16, or
0.4375
If we identify:
P = genotype frequency of AA
Q = genotype frequency of Aa
R = genotype frequency of aa
Then p = gene frequency of A = P + ½ Q
and q = gene frequency of a = R + ½ Q
…. and p + q = 1.
An elementary population genetics model has four steps.
In the absence of selection, genotype
frequencies go to the Hardy-Weinberg
equilibrium.
p2 + 2pq +q2 = 1
Hardy-Weinberg
frequencies of
genotypes AA, Aa,
and aa in relation
to the frequency of
the gene a.
We can test, by
observation,
whether
genotypes in a
population are
in H-W
equilibrium.
We don’t really expect to see genotypes in natural
populations in H-W equilibrium, mainly because of
natural selection.
A simple model of selection can concentrate on how
genotypic frequencies are modified between birth
and the adult reproductive stage.
Start with a simple situation…..
Natural selection operating on a single genetic
locus, at which there are two alleles, A and a, with A
dominant. Suppose that the three possible
genotypes have the following relative chances of
survival from birth to the adult stage:
Genotype
AA, Aa
Chance of
survival
1
Aa
1-s
“s” is a number between 0 and 1, and is called the
selection coefficient.
With the fitnesses given, selection will act to eliminate the
a gene and fix the A allele. To “fix” a gene means to carry
its frequency up to one.
How rapidly will the population change through time?
We need an expression of the gene frequency of A (p') in
one generation in terms of its frequency in the previous
generation (p). The difference between the two:
Δp = p‘ – p
Is the change in gene frequency between two successive
generations.
Look at algebraic example:
Genotype
AA
Aa
Aa
frequency
p2
2pq
q2
fitness
1
1
1–s
2pq
q2(1 – s)
Birth
Adult
relative freq. p2
frequency
p2/1 – sq2 2pq/1 –
sq2
q2(1 – s)/1 –
sq2
And a numerical example
AA
Aa
aa
Birth
Total
Number
1
18
81
Frequency
0.01
0.18
0.81
Fitness
1
1
0.9
Number
1
18
73
Frequency
1/92
18/92
73/92
100
Adult
92
So, how are p’ and p2 related?
Recall that the frequency of A at any time is equal to the
frequency of AA plus half the frequency of Aa. So:
p 2  pq
p
p 

2
2
1  sq
1  sq
Since the denominator is less than one, p΄ is greater than
p. Selection is increasing the frequency of A.
What is Δp?
p
p  p  p 
2
1  sq
p  p  spq

2
1  sq
2
spq

2
1  sq
2
We can use this to calculate the change in gene
frequency between successive generations for any
selection coefficient, s, and any gene frequencies.
Similarly, if we know the gene frequencies for two
successive generations we can rearrange the equation
to solve for the selection coefficient:
p
s
pq 2
So, how rapidly can selection change gene frequencies?
Apply the model to Biston betularia
If we can come up with estimates of the frequencies of the
different color forms at two different times, we can use
them to estimate the gene frequencies, and substitute
those into our equation to get an estimate of s.
A disadvantageous gene in mutation-selection balance has
a frequency of m/s, where m is the mutation rate and s is its
selective advantage. J.B.S. Haldane guessed that m = 10-6
and s = 0.1 for the melanic C gene. This would give the
gene a frequency of 10-5 in 1848 when it was first seen. By
1898, the frequency of the light-colored genotype was 110% in polluted areas, give it a frequency of about 0.2, and
the C gene a frequency of about 0.8
So, what selection coefficient would generate an
increase in the frequency of the C allele from 10-5 to 0.8
in 50 generations? If our equation is iterated 50 times,
we find that a selection coefficient of about 0.33 will lead
to the observed change in gene frequencies. In other
words, the peppered moths would have about 2/3 the
survival rate of the melanic moths.
Theoretical changes in gene frequencies in
the evolution of melanism in the peppered
moth
We can get a second estimate of fitness by looking at the
survivorship of different genotypes in mark-recapture studies