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Neutral theory 1: Genetic load and introduction Neutral theory 1. Mutation 2. Polymorphism 3. Substitution
Neutral theory: connected these is a new (radical) way
1 Neo-Darwinism 1. genetic variation arises at random via mutation - - PDF document
Neutral theory 1: Genetic load and introduction Neutral theory 1. Mutation 2. Polymorphism Neutral theory: connected these is a new (radical) way 3. Substitution 1 Neo-Darwinism 1. genetic variation arises at random via mutation and
Neutral theory: connected these is a new (radical) way
purifying selection in micro-evolution
morphological variation.
heterozygous at most loci
balancing selection are rare with respect to purifying selection in micro-evolution
variation to connect with morph. variation
homozygous at most loci
“It is altogether unlikely that two genes would have identical selective values under all the conditions under which they may coexist in a population. … cases of neutral polymorphism do not exist … it appears probable that random fixation is of negligible evolutionary importance” ⎯Ernst Mayr
(A) Diagram of a protein gel electrophoresis apparatus, and (B) a photograph of a “stained” protein gel, the blue “blotches” are the proteins, their position indicates how far they migrated in the electric field.
Two alleles (A and a) with frequencies p = q = 0.5: Survival to reproduce: AA = 40% Aa = 50% aa = 30% The relative fitness values are: AA = 0.8 Aa = 1 aa = 0.6 The mean fitness of the population = 0.25(0.8) + 0.5(1) + 0.25(0.6) = 0.85 The load of this population (L) = 1 – 0.85 = 0.15
[Note that if every member of the population had the same genotype the average fitnes would equal 1 and the load on the population would be zero.]
Population declines: Genetic death > reproductive excess
20 40 60 80 100 120
1 2 Background mortality when all individuals have the same fitness
no selective death: large excess Background mortality when all individuals have the same fitness no selective death: small excess 20 40 60 80 100 120 1 2
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 26 51 76 101 126 151 176 201 226 251 Generations Frequency of a allele s = 0.1 s = 0.5 s = 0.9 s = 0 s = 0.01 Change in recessive allele frequency over time under different intensities of negative selection
Consider a new mutation to an beneficial dominant allele: it takes time for selection to remove the “old” [deleterious recessive] allele from the population. There is a cost to selection, in genetic death, during this time period
Population declines: Genetic death > reproductive excess
Assume directional selection of a new mutation: C × Ne gives the total selective death; this must be sum over generations it take to fix the allele
W L
= = survive that proportion selection to due die that propotion C
allele fix the to it takes s generation all
Suppose L = 0.1
Load = 10% population reduction Total size = 500 individuals Reproductive size: 450 Cost of selection (C) = L/ = 0.1/0.9 = 0.111 C1 x Ne = 50 extra individuals per 1st generation Total generation to fix allele = 100 Population 1: Reproductive excess = 0 Generation = 53
W Population 2: Reproductive excess = 0.1 Generation = 100
Population 2: Cost = C - R Cost = 0.111 – 0.1 = 0.011 … “soft selection” 450 49.95 400.05 450 4.95 445.05 400.05 44.40555 355.6445 445.05 4.89555 440.1545 355.6445 39.47653 316.1679 440.1545 4.841699 435.3128 316.1679 35.09464 281.0733 435.3128 4.78844 430.5243 281.0733 31.19913 249.8741 430.5243 4.735767 425.7885 249.8741 27.73603 222.1381 425.7885 4.683674 421.1049 222.1381 24.65733 197.4808 421.1049 4.632154 416.4727 197.4808 21.92037 175.5604 416.4727 4.5812 411.8915 175.5604 19.48721 156.0732 411.8915 4.530807 407.3607 156.0732 17.32413 138.7491 407.3607 4.480968 402.8797 Population 1: Reproductive excess = 0 C = 0.111 Extinction: At gen. 53 N =0 Population 2: Reproductive excess = 0.1 C = 0.111 – 0.1 = 0.011 Fixation: At gen. 100 N = 165
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 26 51 76 101 126 151 176 201 226 251 s = 0.1 s = 0.5 s = 0.9 s = 0 s = 0.01Let’s assume: (i) new mutations are deleterious alleles, and (ii) recessive. Remember the approximation of the equilibrium frequency of deleterious alleles [See population genetics, Topic 8 for a review]:
q = (µ/s)1/2
Remember that population load is:
L = 1 -
And remember that the average fitness under these assumptions was: = 1 – sq2 We can make substitutions:
L = 1 - L = 1 – (1 – sq2) L = 1 – (1 – s(µ/s)) L = 1 – (1 – µ) L = µ
It is interesting that we estimate that the load is equal to the mutation rate. Because it suggests that the load is approximately independent of the reduction in fitness caused by the mutant (s).
W W
W
Frequency of a allele
µ = 0.0001
a
0.2 0.4 0.6 0.8 1 Fitness AA Aa aa Genotypes
wAA > wAa > waa
Type 1: Positive Darwinian selection: directional selection for fixation of a new and beneficial mutation in a population . Positive selection: Same as above. [Note that the above term is also shortened to “Darwinian selection”; this is a bad habit of which I am very guilty.] Type 2: Negative Darwinian selection: directional selection for removal of a new and deleterious mutation from a population. Negative selection: same as “negative Darwinian selection”. Purifying selection: same as negative selection
Deleterious recessive Genotype AA Aa aa Frequency p0
2
2p0q0 q0
2
wmodel
1 1 1 - s
w
1 1 0.66
Very high mutation rate (0.01) results in only a small shift in the long term average allele frequency under overdominant selection drift + selection µ = 0.01 Nes = 1000 drift + mutation + selection
The model Genotype AA Aa aa Frequency p0
2
2p0q0 q0
2
w
1 – s1 1 1 – s2
This load persists! Polymorphism is stable; this load never goes away
The model Genotype AA Aa aa Frequency p0
2
2p0q0 q0
2
w
1 – s1 1 1 – s2
Segregational load is a big problem for the balance school: Well known examples exist; Haemoglobin, MHC locus, etc. Balance school would extend this to most polymorphic loci in the genome. Let’s see if this will work. Humans: 30% of loci are polymorphic (from Harris 1966) 30,000 protein-coding genes (from recent genome projects), so 9000 are polymorphic Let’s assume a very small load on average: L = 0.001 Let’s assume that only half are under balancing selection (4500) [remember the balance school predicted a majority would be under balancing selection] Fitness of an individual locus = 0.999 Fitness over whole genome = 0.9994500 = 0.011 Load = 1- 0.011 = 0.989 [That is huge!!!] Cost = 0.989/0.011 = 89 [Do you know of any humans with families that big?]
Bitter tasting Tasty mimics (Papilio)