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-> = 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: 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? - and qeq = v/ v + μ

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? - and qeq = v/ v + μ = 0.00008/0.00009 = 0.89

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? - and qeq = v/ v + μ = 0.00008/0.00009 = 0.89 - 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 mutations.

- 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:

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.