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Population Genetics. Macrophage. CCR5 CCR5- D 32. What accounts for this variation? Random? Past epidemics (plague, smallpox)?. What will happen to this variation in the future? Will D 32 allele increase in frequency?. - PowerPoint PPT Presentation
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Population Genetics
Macrophage
CCR5 CCR5-32
What accounts for this variation? Random? Past
epidemics (plague, smallpox)?
What will happen to this variation in the future? Will 32 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-W1) 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-WIf you know the allele frequencies, you can
predict the genotype frequencies:
Q: In S. France, the frequency of the 32 allele is 10% (i.e., q=0.10). What proportion of individuals will be homozygous for the allele? What proportion will be heterozygous?
AA Aa aa
p2 2pq q2
Usefulness of H-WIf 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 alleleq = freq of B alleler = freq of O allele
Expansion of [p + q + r]2 = p2 + q2 + r2+ 2pq + 2pr + 2qr
Assumptions of H-W1) 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?SelectionMutationNon-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?
MM MN NNp2 2pq q2
0.294 0.496 0.209
Genotypes Expected Observed MM 294.3 298 MN 496.4 489 NN 209.3 213
p=.5425 q=.4575
Test for H-W Genotype Frequencies
Genotypes Expected Observed MM 294.3 298 MN 496.4 489 NN 209.3 213
Importance of H-WH-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. Selection2. Mutation3. Genetic Drift4. Nonrandom Mating5. 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
25100% = 1
50100% = 1
2080% = 0.8
Gamete contribution
25/95 A 25/95 A; 25/95 a
20/95 a
New allele frequencies?
New genotype frequencies (assume random mating):
p = = ____ q = = ____
AA Aa aa0.28 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 frequencyGenotype 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(32)=0.1Genotype frequency:
+/+ p2=0.81
+/32 2pq=0.18
32/32 q2=0.01
Relative Fitness W+/+=0.99 W+/32=0.99 W32/32=1.0Average fitness W = 0. 81*0.99 + 0.18*0.99 + 0.01*1 = 0.9901
q'=q2W32/32/W + pqW+/32 /W=0.01009 +0.089991=0.100091
p’= 1-q’ = 0.89999
Next generation genotype freq.
p2
0.809982pq
0.18016q2
0.01002
q q2
Selection will increase the frequency of 32 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 advantageFrequency dependence
Selection varying in space or time
Heterozygote advantage
Fitness A a a aA A
HbA/HbA HbA/HbS HbS/HbS
Relative Fitness 0.88 1.0 0.14
Fitness (in symbols) 1-t 1 1-s
Selection coefficients t=0.12 s=0.86
Relative fitness of hemoglobin genotypes in Yorubans
Equilibrium frequencies:peq = s/(s+t) = 0.86/(0.12+0.86) = 0.88qeq = t/(s+t) = 0.12/(0.12+0.86) = 0.12Predict the genotype frequencies (at birth):HW proportions 0.774 0.211 0.0144
Variable selection: genotypes have different fitness effects in different environments
0.4
0.5
0.6
0.7
0.8
0.9
1
Env. 1 Env. 2 Env. 3
AAAaaa
Fitness
Frequency-dependent selection
Other Examples of Freq-dep. Selection