Transcript PopGen1_post

Population Genetics
Macrophage
CCR5
CCR5-D32
What accounts for this
variation? Random? Past
epidemics (plague, smallpox)?
What will happen to this variation
in the future? Will D32 allele
increase in frequency?
These are the questions that “population
genetics” is designed to address
Hardy-Weinberg Principle
1. Allele frequencies remain constant
from generation to generation unless
some outside force is acting to change
them
2. When an allele is rare, there are many
more heterozygotes than
homozygotes (if p is small, then ____
is very small)
Assumptions of H-W
1) Mating is random across the entire population.
2) All genotypes have equal viability and fertility (no
selection).
3) Migration into the population can be ignored.
4) Mutation does not occur, or is so rare it can be
ignored.
5) Population is large enough that the allele
frequencies do not change from generation to
generation due to chance (random genetic drift).
6) Allele frequencies are the same in females and
males.
Usefulness of H-W
If you know the allele frequencies, you can
predict the genotype frequencies:
AA
Aa
aa
p2
2pq
q2
Q: In S. France, the frequency of the D32 allele
is 10% (i.e., q=0.10). What proportion of
individuals will be homozygous for the allele?
What proportion will be heterozygous?
Usefulness of H-W
If you know the frequency of one of the
homozygous genotypes, you can estimate
allele frequencies, and predict the frequencies
of the other genotypes.
Q: Among individuals of European descent,
1/1700 newborns have cystic fibrosis (a
recessive genetic disorder). What proportion
of this population are heterozygous carriers?
Hint: q2 = 1/1700 = 0.00059
A:
Multiple Alleles: ABO blood
types
p = freq of A allele
q = freq of B allele
r = freq of O allele
Expansion of [p + q + r]2 =
p2 + q2 + r2+ 2pq + 2pr + 2qr
Assumptions of H-W
1) Mating is random across the entire population.
2) All genotypes have equal viability and fertility (no
selection).
3) Migration into the population can be ignored.
4) Mutation does not occur, or is so rare it can be
ignored.
5) Population is large enough that the allele
frequencies do not change from generation to
generation due to chance (random genetic drift).
6) Allele frequencies are the same in females and
males.
What happens when any of
these assumptions are
violated?
Selection
Mutation
Non-random mating
_______________
_______________
If any of these
processes are
occurring, will tend to
get ____________
from H-W expected
proportions
How can we detect deviation
from H-W expectations?
Do observed genotype frequencies match
HW expectations?
Do observed genotype frequencies match
HW expectations?
q=.4575
p=.5425
MM
p2
0.29 4
Gen otype s
MM
MN
NN
MN
2 pq
0.49 6
Expe ct ed
2 94. 3
4 96. 4
2 09. 3
NN
q2
0.20 9
Ob se rve d
2 98
4 89
2 13
Test for H-W Genotype Frequencies
Gen otype s
MM
MN
NN
Expe ct ed
2 94. 3
4 96. 4
2 09. 3
Ob se rve d
2 98
4 89
2 13
Importance of H-W
H-W is an important tool for population
genetics.
If assumptions are met, we can use it to
estimate allele and genotype
frequencies that would otherwise be
difficult to measure.
If assumptions are not met (can be tested
statistically), then we know that some
outside force is perturbing allele or
genotype frequencies.
Change in allele frequencies
over generations
Evolution is defined as a change
in allele (or genotype) frequencies
over generations, and evolution
will be caused by violation of any
of the assumptions of H-W.
Forces that cause deviation
from H-W (evolution)
1.
2.
3.
4.
5.
Selection
Mutation
Genetic Drift
Nonrandom Mating
Gene Flow (Migration)
Genotype A has a constitutive mutation for enzyme production in
the lactose operon.
B is the normal inducible lactose operon.
A and B grown together in environment with limited lactose.
p = 0.5, q = 0.5
Genotypes
Number:
AA
25
Aa
50
aa
25
Survival to
reproduction
25
100% = 1
50
100% = 1
20
80% = 0.8
Gamete
contribution
25/95 A
25/95 A;
25/95 a
20/95 a
= ____
New allele frequencies? p =
q=
= ____
New genotype frequencies (assume random
mating):
AA
Aa
aa
0.28
0.50
0.22
Consistent differences in survival or
reproduction between genotypes =
genotypic-specific differences in
fitness
When fitness values are expressed
on a scale such that highest
fitness=1, then the values are
called relative fitness
To conveniently calculate change in
allele frequency due to selection,
need concept of average fitness
Change in allele frequency
Genotype
AA
Aa
aa
Genotype Frequency
p2
2pq
q2
Relative Fitness
WAA
WAa
Waa
W=average fitness= (p2WAA)+ (2pqWAa)+ (q2WAa)
Freq of A after one gen. of selection:
p' = p2 WAA/W + pqWAa/W
Freq of a after one gen. of selection: (1-p’) or:
q'= q2 Waa/W + pqWAa/W
CCR5 Example; p(+)=0.9; q(D32)=0.1
Genotype
frequency:
Relative Fitness
+/+
p2=0.81
W+/+=0.99
+/D32
2pq=0.18
W+/D32=0.99
D32/D32
q2=0.01
WD32/D32=1.0
Average fitness W = 0. 81*0.99 + 0.18*0.99 + 0.01*1 = 0.9901
q'=q2WD32/D32/W + pqW+/D32 /W=0.01009 +0.089991=0.100091
p’= 1-q’ = 0.89999
Next generation
genotype freq.
p2
0.80998
2pq
0.18016
q2
0.01002
q
q2
Selection will increase the
frequency of D32 allele
Selection is relatively weak
The favored allele is recessive
and the favored genotype is very rare
The change in allele frequency (response to
selection) will be relatively slow
Response to selection can be
fast!
Selection is
strong
Favored
allele is
partially
dominant
Both alleles
are
common
Selection is not always
“Directional”
Heterozygote advantage
Frequency dependence
Selection varying in space or time
Heterozygote
advantage
Fitnes
s
AA
Aa
aa
Relative fitness of hemoglobin genotypes in Yorubans
Relative Fitness
HbA/HbA
0.88
HbA/HbS
1.0
HbS/HbS
0.14
Fitness (in symbols)
1-t
1
1-s
Selection coefficients
t=0.12
Equilibrium frequencies:
peq = s/(s+t) = 0.86/(0.12+0.86) = 0.88
qeq = t/(s+t) = 0.12/(0.12+0.86) = 0.12
Predict the genotype frequencies (at birth):
HW proportions
0.774
0.211 0.0144
s=0.86
Variable selection: genotypes have different
fitness effects in different environments
1
0.9
0.8
AA
Aa
aa
Fitness 0.7
0.6
0.5
0.4
Env. 1
Env. 2
Env. 3
Frequency-dependent
selection
Other Examples of Freq-dep. Selection