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

2. Deviations from HWE
I. Mutation
A. Basics:

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

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

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 =
'A' will increase by the same amount.

6. 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 =
'A' will increase by the same amount.
4. So, the new gene frequencies will be:
p1 = p + μq =
q1 = q - μq = q(1-μ) =

7. 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) =
q1 = q - μq = q(1-μ) =
5. How about with both FORWARD and backward mutation?
Δq = νp - μq

8. 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) =
q1 = q - μq = q(1-μ) =
5. How about with both FORWARD and backward mutation?
Δq = νp - μq
- so, if A -> a =v = and a-> = μ = ,
and p = 0.6 and q = 0.4, then:

9. 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) =
q1 = q - μq = q(1-μ) =
5. How about with both FORWARD and backward mutation?
Δq = νp - μq
- so, if A -> a =v = and a->A = μ = ,
and p = 0.6 and q = 0.4, then:
Δq = νp - μq = =
q1 = =

10. 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) =
q1 = q - μq = q(1-μ) =
5. How about with both FORWARD and backward mutation?
- and qeq = v/ v + μ

11. 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) =
q1 = q - μq = q(1-μ) =
5. How about with both FORWARD and backward mutation?
- and qeq = v/ v + μ = / = 0.89

12. 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) =
q1 = q - μq = q(1-μ) =
5. How about with both FORWARD and backward mutation?
- and qeq = v/ v + μ = / = 0.89
- Δq = (.11)( ) - (.89)( ) = check.

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

14. 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.

15. 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.

16. - 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

17. - 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

18. - 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

19. - 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)

20. - 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

21. 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

22. 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

23. 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

24. 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 = √ ( ) * ( ) = 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

25. - 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 = √ ( ) * ( ) = 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 = √ ( ) * ( ) = 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 = √ ( ) * ( ) = 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

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

27. 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 F1
.35 .3

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

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

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

31. 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

32. 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).

33. 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 = so F = ( )/.5 = 0.4 AA Aa aa .2 .6
offspring
ALL AA
1/4AA:1/2Aa:1/4aa
ALL aa F1
.35 .3

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

35. 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

36. 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

37. 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....

38. 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.

39. 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....?

40. 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.