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 p2 Aa 2pq aa 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?
p=.5425
MM p2 0.29 4 MN 2 pq 0.49 6
q=.4575
NN q2 0.20 9
Gen otype s MM MN NN
Expect ed 2 94. 3 4 96. 4 2 09. 3
Ob se rve d 2 98 4 89 2 13
Test for H-W Genotype Frequencies
Gen otype s MM MN NN Expect 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
Gamete contribution
25 100% = 1
25/95 A
50 100% = 1
25/95 A; 25/95 a
20 80% = 0.8
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
Relative Fitness
p2
WAA
2pq
WAa
q2
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: +/+ p2=0.81 W+/+=0.99 +/D32 2pq=0.18
W+/D32=0.99
Relative Fitness
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
HbA/HbA 0.88 1-t HbA/HbS 1.0 1 HbS/HbS 0.14 1-s
Relative Fitness Fitness (in symbols)
Selection coefficients
t=0.12
s=0.86
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
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