III. Modeling Selection A. Selection for a Dominant Allele B. Selection for an Incompletely Dominant Allele C. Selection that Maintains Variation

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote p = 0.4, q = 0.6 AA Aa aa Parental "zygotes" 0.16 0.48 0.36 = 1.00 prob. of survival (fitness) 0.4 0.8 0.2 Relative Fitness 0.5 (1-s) 1 0.25 (1-t) Survival to Reproduction 0.08 0.09 = 0.65 Geno. Freq., breeders 0.12 0.74 0.14 Gene Freq's, gene pool p = 0.49 q = 0.51 Genotypes, F1 0.24 0.50 0.26 = 100

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote - Consider an 'A" allele. It's probability of being lost from the population is a function of:

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote - Consider an 'A" allele. It's probability of being lost from the population is a function of: 1) probability it meets another 'A' (p)

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote - Consider an 'A" allele. It's probability of being lost from the population is a function of: 1) probability it meets another 'A' (p) 2) rate at which these AA are lost (s).

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote - Consider an 'A" allele. It's probability of being lost from the population is a function of: 1) probability it meets another 'A' (p) 2) rate at which these AA are lost (s). - So, prob of losing an 'A' allele = ps

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote - Consider an 'A" allele. It's probability of being lost from the population is a function of: 1) probability it meets another 'A' (p) 2) rate at which these AA are lost (s). - So, prob of losing an 'A' allele = ps - Likewise the probability of losing an 'a' = qt

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote - Consider an 'A" allele. It's probability of being lost from the population is a function of: 1) probability it meets another 'A' (p) 2) rate at which these AA are lost (s). - So, prob of losing an 'A' allele = ps - Likewise the probability of losing an 'a' = qt - An equilibrium will occur, when the probability of losing A an a are equal; when ps = qt.

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote - An equilibrium will occur, when the probability of losing A an a are equal; when ps = qt.

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote - An equilibrium will occur, when the probability of losing A an a are equal; when ps = qt. - substituting (1-p) for q, ps = (1-p)t ps = t - pt ps +pt = t p(s + t) = t peq = t/(s + t)

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote - An equilibrium will occur, when the probability of losing A an a are equal; when ps = qt. - substituting (1-p) for q, ps = (1-p)t ps = t - pt ps +pt = t p(s + t) = t peq = t/(s + t) - So, for our example, t = 0.75, s = 0.5 - so, peq = .75/1.25 = 0.6

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote - so, peq = .75/1.25 = 0.6 p = 0.6, q = 0.4 AA Aa aa Parental "zygotes" 0.36 0.48 0.16 = 1.00 prob. of survival (fitness) 0.4 0.8 0.2 Relative Fitness 0.5 (1-s) 1 0.25 (1-t) Survival to Reproduction 0.18 0.04 = 0.70 Geno. Freq., breeders 0.26 0.68 0.06 Gene Freq's, gene pool p = 0.6 q = 0.4 CHECK Genotypes, F1 = 100

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote - so, peq = .75/1.25 = 0.6 - so, if p > 0.6, it should decline to this peq p = 0.7, q = 0.3 AA Aa aa Parental "zygotes" 0.49 0.42 0.09 = 1.00 prob. of survival (fitness) 0.4 0.8 0.2 Relative Fitness 0.5 (1-s) 1 0.25 (1-t) Survival to Reproduction 0.25 0.48 0.02 = 0.75 Geno. Freq., breeders 0.33 0.64 0.03 Gene Freq's, gene pool p = 0.65 q = 0.35 CHECK Genotypes, F1 0.46 0.12 = 100

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote - so, peq = .75/1.25 = 0.6 - so, if p > 0.6, it should decline to this peq 0.6

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote 2. Multiple Niche Polymorphism -

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote 2. Multiple Niche Polymorphism - - equilibrium can occur if AA and aa are each fit in a given niche, within the population. The equilibrium will depend on the relative frequencies of the niches and the selection differentials...

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote 2. Multiple Niche Polymorphism - - equilibrium can occur if AA and aa are each fit in a given niche, within the population. The equilibrium will depend on the relative frequencies of the niches and the selection differentials... - can you think of an example??

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote 2. Multiple Niche Polymorphism - - equilibrium can occur if AA and aa are each fit in a given niche, within the population. The equilibrium will depend on the relative frequencies of the niches and the selection differentials... - can you think of an example?? Papilio butterflies... females mimic different models and an equilibrium is maintained; in fact, an equilibrium at each locus, which are also maintained in linkage disequilibrium.

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote 2. Multiple Niche Polymorphism 3. Frequency Dependent Selection

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote 2. Multiple Niche Polymorphism 3. Frequency Dependent Selection - the fitness depends on the frequency...

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote 2. Multiple Niche Polymorphism 3. Frequency Dependent Selection - the fitness depends on the frequency... - as a gene becomes rare, it becomes advantageous and is maintained in the population...

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote 2. Multiple Niche Polymorphism 3. Frequency Dependent Selection - the fitness depends on the frequency... - as a gene becomes rare, it becomes advantageous and is maintained in the population... - "Rare mate" phenomenon...

Elderflower orchids: - don’t produce nectar - bumblebees visit most common flower color and get discouraged, try the other color…. Back and forth. - visit equal NUMBERS of the two colors, but that means that a plant with the rare color is visited more often.

(of yellow flowers)

- Morphs of Heliconius melpomene and H. erato Mullerian complex between two distasteful species... positive frequency dependence in both populations to look like the most abundant morph

C. Selection that Maintains Variation 1. Heterosis - selection for the heterozygote 2. Multiple Niche Polymorphism 3. Frequency Dependent Selection 4. Selection Against the Heterozygote p = 0.4, q = 0.6 AA Aa aa Parental "zygotes" 0.16 0.48 0.36 = 1.00 prob. of survival (fitness) 0.8 0.4 0.6 Relative Fitness 1 0.5 0.75 Corrected Fitness 1 + 0.5 1.0 1 + 0.25 formulae 1 + s 1 + t

4. Selection Against the Heterozygote p = 0.4, q = 0.6 AA Aa aa Parental "zygotes" 0.16 0.48 0.36 = 1.00 prob. of survival (fitness) 0.8 0.4 0.6 Relative Fitness 1 0.5 0.75 Corrected Fitness 1 + 0.5 1.0 1 + 0.25 formulae 1 + s 1 + t

4. Selection Against the Heterozygote - peq = t/(s + t) p = 0.4, q = 0.6 AA Aa aa Parental "zygotes" 0.16 0.48 0.36 = 1.00 prob. of survival (fitness) 0.8 0.4 0.6 Relative Fitness 1 0.5 0.75 Corrected Fitness 1 + 0.5 1.0 1 + 0.25 formulae 1 + s 1 + t

4. Selection Against the Heterozygote - peq = t/(s + t) - here = .25/(.50 + .25) = .33 p = 0.4, q = 0.6 AA Aa aa Parental "zygotes" 0.16 0.48 0.36 = 1.00 prob. of survival (fitness) 0.8 0.4 0.6 Relative Fitness 1 0.5 0.75 Corrected Fitness 1 + 0.5 1.0 1 + 0.25 formulae 1 + s 1 + t

4. Selection Against the Heterozygote - peq = t/(s + t) - here = .25/(.50 + .25) = .33 - if p > 0.33, then it will keep increasing to fixation. p = 0.4, q = 0.6 AA Aa aa Parental "zygotes" 0.16 0.48 0.36 = 1.00 prob. of survival (fitness) 0.8 0.4 0.6 Relative Fitness 1 0.5 0.75 Corrected Fitness 1 + 0.5 1.0 1 + 0.25 formulae 1 + s 1 + t

4. Selection Against the Heterozygote - peq = t/(s + t) - here = .25/(.50 + .25) = .33 - if p > 0.33, then it will keep increasing to fixation. - However, if p < 0.33, then p will decline to zero... AND THERE WILL BE FIXATION FOR A SUBOPTIMAL ALLELE....'a'... !! UNSTABLE EQUILIBRIUM!!!!

Population Genetics I. Basic Principles II. X-linked Genes III. Modeling Selection IV. OTHER DEVIATIONS FROM HWE

Deviations from HWE I. Mutation A. Basics:

Deviations from HWE I. Mutation A. Basics: 1. Consider a population with: f(A) = p = .6 f(a) = q = .4

Deviations from HWE I. Mutation A. Basics: 1. Consider a population with: f(A) = p = .6 f(a) = q = .4 2. Suppose 'a' mutates to 'A' at a realistic rate of: μ = 1 x 10-5

Deviations from HWE I. Mutation A. Basics: 1. Consider a population with: f(A) = p = .6 f(a) = q = .4 2. Suppose 'a' mutates to 'A' at a realistic rate of: μ = 1 x 10-5 3. Well, what fraction of alleles will change? 'a' will decline by: qm = .4 x 0.00001 = 0.000004 'A' will increase by the same amount.

Deviations from HWE I. Mutation A. Basics: 1. Consider a population with: f(A) = p = .6 f(a) = q = .4 2. Suppose 'a' mutates to 'A' at a realistic rate of: μ = 1 x 10-5 3. Well, what fraction of alleles will change? 'a' will decline by: qm = .4 x 0.00001 = 0.000004 'A' will increase by the same amount. 4. So, the new gene frequencies will be: p1 = p + μq = .600004 q1 = q - μq = q(1-μ) = .399996

Deviations from HWE I. Mutation A. Basics: 4. So, the new gene frequencies will be: p1 = p + μq = 1 - q + μq = 1- q(1-μ) = .600004 q1 = q - μq = q(1-μ) = .399996 5. How about with both FORWARD and backward mutation? Δq = νp - μq

Deviations from HWE I. Mutation A. Basics: 4. So, the new gene frequencies will be: p1 = p + μq = 1 - q + μq = 1- q(1-μ) = .600004 q1 = q - μq = q(1-μ) = .399996 5. How about with both FORWARD and backward mutation? Δq = νp - μq - so, if A -> a =v = 0.00008 and a->A = μ = 0.00001, and p = 0.6 and q = 0.4, then:

Deviations from HWE I. Mutation A. Basics: 4. So, the new gene frequencies will be: p1 = p + μq = 1 - q + μq = 1- q(1-μ) = .600004 q1 = q - μq = q(1-μ) = .399996 5. How about with both FORWARD and backward mutation? Δq = νp - μq - so, if A -> a =v = 0.00008 and a->A = μ = 0.00001, and p = 0.6 and q = 0.4, then: Δq = νp - μq = 0.000048 - 0.000004 = 0.000044 q1 = .4 + 0.000044 = 0.400044

Deviations from HWE I. Mutation A. Basics: 5. How about with both FORWARD and backward mutation? - Δq = νp - μq - and qeq = v/ v + μ

Deviations from HWE I. Mutation A. Basics: 5. How about with both FORWARD and backward mutation? - Δq = νp - μq - and qeq = v/ v + μ - and qeq = v/ v + μ = 0.00008/0.00009 = 0.89

Deviations from HWE I. Mutation A. Basics: 5. How about with both FORWARD and backward mutation? - Δq = νp - μq - and qeq = v/ v + μ - and qeq = v/ v + μ = 0.00008/0.00009 = 0.89 - so, if Δq = νp – μq, then: Δq = (.11)(0.00008) - (.89)(0.00001) = 0.0..... check.

Deviations from HWE I. Mutation A. Basics: B. Other Considerations:

B. Other Considerations: Deviations from HWE I. Mutation A. Basics: B. Other Considerations: - Selection: Selection can BALANCE mutation... so a deleterious allele might not accumulate as rapidly as mutation would predict, because it it eliminated from the population by selection each generation. We'll model these effects later.

B. Other Considerations: Deviations from HWE I. Mutation A. Basics: B. Other Considerations: - Selection: Selection can BALANCE mutation... so a deleterious allele might not accumulate as rapidly as mutation would predict, because it it eliminated from the population by selection each generation. We'll model these effects later. - Drift: The probability that a new allele (produced by mutation) becomes fixed (q = 1.0) in a population = 1/2N (basically, it's frequency in that population of diploids). In a small population, this chance becomes measureable and likely. So, NEUTRAL mutations have a reasonable change of becoming fixed in small populations... and then replaced by new mutation

- Consider two populations: Deviations from HWE I. Mutation II. Migration A. Basics: - Consider two populations: p2 = 0.7 q2 = 0.3 p1 = 0.2 q1 = 0.8

- Consider two populations: Deviations from HWE I. Mutation II. Migration A. Basics: - Consider two populations: p2 = 0.7 q2 = 0.3 p1 = 0.2 q1 = 0.8 suppose migrants immigrate at a rate such that the new immigrants represent 10% of the new population

- Consider two populations: Deviations from HWE I. Mutation II. Migration A. Basics: - Consider two populations: p2 = 0.7 q2 = 0.3 p1 = 0.2 q1 = 0.8 suppose migrants immigrate at a rate such that the new immigrants represent 10% of the new population

- Consider two populations: Deviations from HWE I. Mutation II. Migration A. Basics: - Consider two populations: p2 = 0.7 q2 = 0.3 p1 = 0.2 q1 = 0.8 suppose migrants immigrate at a rate such that the new immigrants represent 10% of the new population p(new) = p1(1-m) + p2(m)

- Consider two populations: Deviations from HWE I. Mutation II. Migration A. Basics: - Consider two populations: p2 = 0.7 q2 = 0.3 p1 = 0.2 q1 = 0.8 suppose migrants immigrate at a rate such that the new immigrants represent 10% of the new population p(new) = p1(1-m) + p2(m) p(new) = 0.2(0.9) + 0.7(0.1) = 0.25

Deviations from HWE I. Mutation II. Migration A. Basics: B. Advanced: - Consider three populations: p1 = 0.7 q1 = 0.3 p2 = 0.2 q2 = 0.8 p3 = 0.6 q3 = 0.4

Deviations from HWE I. Mutation II. Migration A. Basics: B. Advanced: - Consider three populations: - How different are they, genetically? (this can give us a handle on how much migration there may be between them...) p1 = 0.7 q1 = 0.3 p2 = 0.2 q2 = 0.8 p3 = 0.6 q3 = 0.4

Deviations from HWE I. Mutation II. Migration A. Basics: B. Advanced: - Consider three populations: - How different are they, genetically? (this can give us a handle on how much migration there may be between them...) - Compute Nei's Genetic Distance: D = -ln [ ∑pi1pi2/ √ ∑pi12 ∑ pi22] p1 = 0.7 q1 = 0.3 p2 = 0.2 q2 = 0.8 p3 = 0.6 q3 = 0.4

Deviations from HWE I. Mutation II. Migration A. Basics: B. Advanced: - Consider three populations: - How different are they, genetically? (this can give us a handle on how much migration there may be between them...) - Compute Nei's Genetic distance: D = -ln [ ∑pi1pi2/ √ ∑pi12 ∑ pi22] - So, for Population 1 and 2: - ∑pi1pi2 = (0.7*0.2) + (0.3*0.8) = 0.38 - denominator = √ (.49+.09) * (.04+.64) = 0.628 D12 = -ln (0.38/0.62) = 0.50 p1 = 0.7 q1 = 0.3 p2 = 0.2 q2 = 0.8 p3 = 0.6 q3 = 0.4

- Compute Nei's Genetic distance: D = -ln [ ∑pi1pi2/ √ ∑pi12 ∑ pi22] - So, for Population 1 and 2: - ∑pi1pi2 = (0.7*0.2) + (0.3*0.8) = 0.38 - denominator = √ (.49+.09) * (.04+.64) = 0.628 D12 = -ln (0.38/0.628) = 0.50 - For Population 1 and 3: - ∑pi1pi2 = (0.7*0.6) + (0.3*0.4) = 0.54 - denominator = √ (.49+.09) * (.36+.16) = 0.55 D13 = -ln (0.54/0.55) = 0.02 - For Population 2 and 3: - ∑pi1pi2 = (0.2*0.6) + (0.8*0.4) = 0.44 - denominator = √ (.04+.64) * (.36+.16) = 0.61 D23 = -ln (0.44/0.61) = 0.33 p1 = 0.7 q1 = 0.3 p2 = 0.2 q2 = 0.8 p3 = 0.6 q3 = 0.4

Deviations from HWE I. Mutation II. Migration III. Non-Random Mating A. Positive Assortative Mating "like phenotype mates with like phenotype"

A. Positive Assortative Mating Deviations from HWE I. Mutation II. Migration III. Non-Random Mating A. Positive Assortative Mating "like phenotype mates with like phenotype" 1. Pattern: AA Aa aa .2 .6 offspring ALL AA 1/4AA:1/2Aa:1/4aa ALL aa .15 + .3 + .15 F1 .35 .3

1. Pattern: 2. Effect: AA Aa aa .2 .6 offspring ALL AA .15 + .3 + .15 F1 .35 .3 2. Effect: - reduction in heterozygosity at this locus; increase in homozygosity.

Deviations from HWE I. Mutation II. Migration III. Non-Random Mating A. Positive Assortative Mating B. Inbreeding 1. Overview:

Deviations from HWE I. Mutation II. Migration III. Non-Random Mating A. Positive Assortative Mating B. Inbreeding 1. Overview: Inbreeding is “like mating with like”, but across the entire genome.

B. Inbreeding 1. Overview: - Autozygous - inherited alleles common by descent - F = inbreeding coefficient = prob. of autozygosity - so, (1-F) = prob. of allozygosity

B. Inbreeding 1. Overview: - Autozygous - inherited alleles common by descent - F = inbreeding coefficient = prob. of autozygosity - so, (1-F) = prob. of allozygosity - SO: f(AA) = p2(1-F) + p2(F) = D f(Aa) = 2pq(1-F) = H (observed) f(aa) = q2(1-F) + q2(F) = R

B. Inbreeding 1. Overview: - Autozygous - inherited alleles common by descent - F = inbreeding coefficient = prob. of autozygosity - so, (1-F) = prob. of allozygosity - SO: f(AA) = p2(1-F) + p2(F) = D f(Aa) = 2pq(1-F) = H (observed) f(aa) = q2(1-F) + q2(F) = R - SO!! the net effect is a decrease in heterozygosity at a factor of (1-F) each generation. - So, the fractional demise of heterozygosity compared to HWE expectations is also a direct measure of inbreeding! F = (2pq - H)/2pq = (Hexp - Hobs)/ Hexp When this is done on multiple loci, the values should all be similar (as inbreeding affects the whole genotype).

B. Inbreeding 1. Overview: - Example: F = (2pq - H)/2pq = (Hexp - Hobs)/ Hexp p = .5, q = .5, expected HWE heterozygosity = 2pq = 0.5 OBSERVED in F1 = 0.3... so F = (.5 - .3)/.5 = 0.4 AA Aa aa .2 .6 offspring ALL AA 1/4AA:1/2Aa:1/4aa ALL aa .15 + .3 + .15 F1 .35 .3

Deviations from HWE I. Mutation II. Migration III. Non-Random Mating A. Positive Assortative Mating B. Inbreeding 1. Overview: 2. Effects:

A. Positive Assortative Mating B. Inbreeding 1. Overview: 2. Effects: Deviations from HWE I. Mutation II. Migration III. Non-Random Mating A. Positive Assortative Mating B. Inbreeding 1. Overview: 2. Effects: - reduce heterozygosity across entire genome

A. Positive Assortative Mating B. Inbreeding 1. Overview: 2. Effects: Deviations from HWE I. Mutation II. Migration III. Non-Random Mating A. Positive Assortative Mating B. Inbreeding 1. Overview: 2. Effects: - reduce heterozygosity across entire genome - rate dependent upon degree of relatedness

A. Positive Assortative Mating B. Inbreeding 1. Overview: 2. Effects: Deviations from HWE I. Mutation II. Migration III. Non-Random Mating A. Positive Assortative Mating B. Inbreeding 1. Overview: 2. Effects: - reduce heterozygosity across entire genome - rate dependent upon degree of relatedness - change in genotypic frequencies but no change in gene frequencies as a result of non-random mating ALONE....

A. Positive Assortative Mating B. Inbreeding 1. Overview: 2. Effects: Deviations from HWE I. Mutation II. Migration III. Non-Random Mating A. Positive Assortative Mating B. Inbreeding 1. Overview: 2. Effects: - reduce heterozygosity across entire genome - rate dependent upon degree of relatedness - change in genotypic frequencies but no change in gene frequencies as a result of non-random mating ALONE.... - BUT... increasing homozygosity may reveal deleterious recessives.

A. Positive Assortative Mating B. Inbreeding 1. Overview: 2. Effects: Deviations from HWE I. Mutation II. Migration III. Non-Random Mating A. Positive Assortative Mating B. Inbreeding 1. Overview: 2. Effects: - reduce heterozygosity across entire genome - rate dependent upon degree of relatedness - change in genotypic frequencies but no change in gene frequencies as a result of non-random mating ALONE.... - BUT... increasing homozygosity may reveal deleterious recessives. - these will be quickly selected against....?

A. Positive Assortative Mating B. Inbreeding 1. Overview: 2. Effects: Deviations from HWE I. Mutation II. Migration III. Non-Random Mating A. Positive Assortative Mating B. Inbreeding 1. Overview: 2. Effects: - reduce heterozygosity across entire genome - rate dependent upon degree of relatedness - change in genotypic frequencies but no change in gene frequencies as a result of non-random mating ALONE.... - BUT... increasing homozygosity may reveal deleterious recessives. - these will be quickly selected against, but that reduces fecundity (inbreeding depression) and reduces genetic variation.