January 25th, 2010
Bioe 109
Winter 2010
Lecture 8
Microevolution 1 - selection
The Hardy-Weinberg-Castle Equilibrium
- consider a single locus with two alleles A1 and A2.
- three genotypes are thus possible: A1A1, A1A2, A2A2.
- let p = frequency of A1 allele
- let q = frequency of A2 allele
- since only two alleles present, p + q = 1
Question: If mating occurs at random in the population, what will the frequencies
of A1 and A2be in the next generation?
- what does “random mating” mean?
- very simply, that matings take place in the population independently of the genotype of an individual.
- if mating is random the frequencies of matings among different genotypes are determined simply by their frequencies.
- random mating can be described among genotypes or, more simply, among the gametes produced by these individuals.
- what are the probabilities of such matings at the gamete level?
- these will occur at proportions determined by the frequencies of the A1 and A2 alleles in the population:
EggSpermZygote Probability
A1x A1 A1A1 p x p = p2
A1x A2 A1A2 p x q = pq
= 2pq
A2x A1 A2A1 q x p = qp
A2x A2 A2A2 q x q = q2
- therefore, zygotes are produced in the following proportions:
A1A1A1A2A2A2
p2 2pq q2
- what are the allele frequencies in the next generation?
1. Frequency of A1= p2 + ½ (2pq)
= p2 + pq
= p(p + q)
= p
2. Frequency of A2= q2 + ½ (2pq)
= q2 + pq
= q(q + p)
= q
NOTE THAT THE FREQUENCIES OF THE ALLELES DO NOT CHANGE
There are three conclusions to be drawn from the H-W principle:
1. Allele frequencies will not change from generation to generation.
2. Genotype proportions will occur according to the “square law”.
- for two alleles (p + q)2 = p2 + 2pq + q2
- for three alleles ( p + q + r)2 = p2 + q2 + r2 + 2pq + 2pr +2qr
3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies
- proportions of genotypes will shift in accordance with those predicted by the square law.
Allele frequenciesGenotype frequencies
p = 0.80, q = 0.20 A1A1 = 0.64; A1A2 = 0.32; A2A2 = 0.04
p = 0.50, q = 0.50 A1A1 = 0.25; A1A2 = 0.50; A2A2 = 0.25
p = 0.20, q = 0.80 A1A1 = 0.04; A1A2 = 0.32; A2A2 = 0.64
Assumptions of Hardy-Weinberg Equilibrium:
1. Random mating
2. Infinite population size (i.e., no genetic drift)
3. No migration (i.e., no gene flow)
4. No mutation
5. No selection
- The H-W principle thus states that allele frequencies in populations will not change unless some evolutionary process is acting to cause a change in allele frequency.
- therefore, this principle thus predicts that no evolution will occur unless one of the above assumptions is violated.
- therefore, the study of microevolution is largely concerned with understanding the conditions under which the above assumptions are violated.
- in other words, it involves determining the relative importance of random drift, migration, mutation, and natural selection in affecting the frequency of genetic polymorphism in natural populations.
- given that many (most?) of these assumptions are likely to violated, an important question to ask is whether Hardy-Weinberg equilibrium ever exists in nature?
Example: Atlantic cod (Gadus morhua) in Nova Scotia
- a sample of 364 fish scored for a DNA SNP (single nucleotide polymorphism)
- the following genotypes were observed.
A1A1 = 109
A1A2 = 182
A2A2 = 73
364
Question: Is this population in Hardy-Weinberg equilibrium?
Testing for Hardy-Weinberg equilibrium
Step 1: Estimate genotype frequencies.
Frequency of A1A1 = 109/364 = 0.2995
Frequency of A1A2 = 182/364 = 0.5000
Frequency of A2A2 = 73/364 = 0.2005
Step 2: Estimate allele frequencies.
Frequency of A1 = p = Freq (A1A1) + ½ Freq (A1A2)
= 0.2995 + ½ (0.5000)
= 0.5495
Frequency of A2 = q= Freq (A2A2) + ½ Freq (A1A2)
= 0.2005 + ½ (0.5000)
= 0.4505
Check that p + q = 0.5495 + 0.4505 = 1
Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium.
- here, we are testing if genotypes are present in proportions consistent with the square law.
Expected No. of A1A1 = p2 x N
= (0.5495)2 x 364
= 109.9
Expected No. of A1A2 = 2pq x N
= 2(0.5495)(0.4505) x 364
= 180.2
Expected No. of A2A2 = q2 x N
= (0.4505)2 x 364
= 73.9
Step 4: Compare observed and expected numbers of genotypes .
GenotypeObservedExpected
A1A1 109 109.9
A1A2 182 180.2
A2A2 73 73.9
- we can test if the proportions differ significantly by performing a Chi-square test (see Box 6.5 on pages 192-193): 2 = (Obs. – Exp.)2 = 0.036
Exp.
- the observed Chi-square is not significant.
- although there is an excellent agreement between observed proportions and those expected under the assumption of Hardy-Weinberg equilibrium.
- does this mean that one, or more, assumptions are in fact being violated.
- if they are, then their effect is so slight as to not cause any appreciable perturbation.
- suppose we were to sample a “mixed population”.
Population APopulation B
p = A1 = 1.0p = A1 = 0
q = A2 = 0q = A2 = 1.0
100 % A1A1100 % A2A2
Mixed population
50% from population A (all A1A1)
50% from population B (all A2A2)
Sample 1000 individuals
ObservedExpected
A1A1 = 500A1A1 = 250
A1A2 = 0A1A2 = 500
A2A2 = 500A2A2 = 250
- here, there is a marked departure from Hardy-Weinberg due to the violation of the random mating assumption.
- a deficiency of heterozygotes always occurs when a “mixed population” is sampled – this is called a Wahlund effect.
A simple model of directional selection
- consider a simple two allele case where p = A1 and q = A2.
- let the relative fitnesses of the three genotypes be:
GenotypeA1A1A1A2A2A2
Rel. fitness w11 w12w22
- there are three equations needed to predict the changes in frequencies of alleles in a population caused by directional selection.
- the first two estimate the fitness of the A1 and A2 alleles:
w1 = p w11 + q w12
w2 = q w22 + p w12
- the mean fitness of the A1 allele is thus a weighted average of its fitnesses in the two genotypes in which it can occur (i.e., A1A1 and A1A2).
- the third equation gives the mean fitness of the entire population:
wbar = pw1 + qw2
- designating the frequency of allele A1 in the next generation as p’
p’ = pw1/ (pw1 + qw2)
= p (w1/wbar)
similarly,q’ = qw2/ (pw1 + qw2)
= q (w2/wbar)
- these equations tell us that changes in the frequencies of alleles “A” and “a” over one generation are proportional to the difference between the allele’s fitness and the mean population fitness
An example: suppose frequencies of alleles A1 and A2 are p = q = 0.50.
GenotypeA1A1A1A2A2A2
Rel. fitnessw11w12w22
10.950.90
- we can intuitively see the outcome - the small a allele will be eliminated from the population.
- how rapidly will this occur?
w1 = p(w11) + q(w12) = 0.5(1.0) + 0.5(0.95) = 0.975
w2 = q(w22) + p(w12) = 0.5(0.9) + 0.5(0.95) = 0.925
wbar = p(w1) + q(w2) = 0.5(0.975) + 0.5(0.925) = 0.950
thus,p’ = p(w1/wbar) = 0.5(0.975/0.95) = 0.513
q’ = q(w2/wbar) = 0.5(0.925/0.95) = 0.487
Conclusion: the frequency of the A1 allele increased by only 0.013 even though the strength of selection is moderately strong.
- even though this is a small change, it will take only 150 generations for the A1 allele to be fixed (i.e., reach a frequency of 1).