1. Given genotype frequencies, calculate allele frequencies in a gene pool !
Alleles = A, a
Genotypes = AA, Aa, aa
Frequency of allele A: f (A) = f (AA) + 1/2 f (Aa)
Frequency of allele a: f (a) = f (aa) + 1/2 f (Aa)
f(A) + f(a) = 1.0
p q = (allele frequencies)
p = 1 - q or, q = 1 - p
p2 + 2pq + q2 = f [AA] + 2 [f(Aa)] + f [aa] = 1 A a A AA Aa a Aa aa
26a

2. Hardy-Weinberg Equilibrium
Parental generation: 2 alleles, r and R
f (R) + f (r) = 1.0
p q = p = 0.1, q = 0.9
In the next generation (F1):
p2 + 2pq + q2 = 1 predicts allele freqs.
F1 genotype Genotype Allele freq.
p2 (.01) RR p = /2= .1
q (.81) rr q = /2 = .9
2pq (.18) Rr
27a-1

3. Hardy-Weinberg Equilibrium
Parental allele frequencies p and q predict F1 generation genotype frequencies, by the formula p2 + 2pq + q2 = 1
Note: parental generation genotype frequencies do NOT
predict F1 generation genotype frequencies!!
27a-2

4. Hardy-Weinberg Equilibrium
Conclusions:
1)Allele frequencies are conserved (i.e., the same) from one generation to the next.
2) genotype frequency reaches Hardy-Weinberg equilibrium in one generation
27a-2

5. Only Hardy-Weinberg’s caveats: But
Hardy-Weinberg Law: allele frequencies in a population
remain constant from generation to generation …..
IF random mating
IF all genotypes are equally viable
IF not disturbed by mutation, selection or whatever
Only
Bottom line: Only in an IDEAL population is genetic diversity conserved forever.

6. Sex = Random sampling of a gene pool
Population of 10 individuals (N = 10)
Phenotypes Red White
Genotypes RR,Rr, rr
Allele frequencies R = 0.6 r = 0.4
Parental gene pool
10 genotypes
20 alleles
Parental gametes
Probability of
F1 r = .4; R =.6
R r r R R R R r r R r
r R R R r R r R R
25A -1

7. F1 genotypes and phenotypes
Genotype Frequency Phenotype Frequency
Rr or rR R_ =.84
rr rr
RR .36
 = 1
 = 1
25A -2

8. If allele frequencies are P and Q in the parental generation, how
do we calculate what they will be in the F1 generation?

9. If allele frequencies are P and Q in the parental generation, how
do we calculate what they will be in the F1 generation?
Genotype frequencies in F1 are calculated by:
p2 + 2pq + q2 = 1
From which we can calculate p and q for F1

10. If the fraction of the population with allele “A” at a given locus
is .7, and the fraction of the population with “a” at the locus is .3,
what will be the expected genotype frequencies in F1, , the next
generation?

11. If the fraction of the population with allele “A” at a given locus
is .7, and the fraction of the population with “a” at the locus is .3,
what will be the expected genotype frequencies in the F1?
Allele frequencies in P (parental generation):
A = .7 = p
a = .3 = q
Expected genotypes and their frequencies in F1:
AA = p2 = .49
aa = q2 = .09
Aa = 2pq = .42

12. What will be the expected phenotypes and their ratios in this
example?

13. What will be the expected phenotypes and their ratios in this
example?
Allele frequencies in P (parental generation):
A = .7 = p
a = .3 = q
Expected genotypes in F1:
AA = p2 = .49
aa = q2 = .09
Aa = 2pq = .42
Expected phenotypes in F1:
A‑ = = .91
aa = .09

14. Heterozygosity defined
H = % heterozygous genotypes for a particular locus
= % heterozygous individuals for a particular locus
= probability that a given individual randomly selected from the population will be heterozygous at a given locus
29f

15. H = ?
Aa aa aa AA aa
aA aa AA aA AA
aa aa AA aa aA
H = 4/15

16. Heterozygosity defined
H (“H-bar”) = average heterozygosity for all loci in a population.
H estimated = % heterozygous loci
those examined
H = 2pq
29f

17. (assuming simple dominance and Hardy-Weinberg Eq.)
Calculating H
(assuming simple dominance and Hardy-Weinberg Eq.)
calculate H if q2 = 0.09
f (a) = 0.3 = q q2 = 0.09
f (A) = 0.7 = p p2 = 0.49
2 pq = 0.42
H = 2pq = (for only 2 alleles)
29a - 1

18. Codominance Genotype Phenotype N AA Red 50 Aa Pink 22 aa White 10
Calculating H
(but if……)
Codominance Genotype Phenotype N
AA Red 50
Aa Pink 22
aa White 10
total = 82
H = 22/82 (don’t need Hardy - Weinberg)
29a - 1

19. Calculating H for 3 alleles: p, q, r
H = 2pq + 2qr + 2pr =
= .58 p q r pp pq p pr qq q pq qr r pr qr rr
29a

20. 1% of golden lion tamarins have diaphragmatic hernias,
a condition expressed only in the homozygous recessive
genotype. Calculate the number of heterozygous individuals
in the wild population (N = 508). Assume Hardy-Weinberg
equilibrium and simple dominance.
Genotypes: AA Aa aa
F1 generation p2 = ? = f (AA)
2pq = ? = f (Aa) = H
q2 = = f (aa)
29ez - 1

21. q2 = .01
q =
= 0.1
p + q = 1 so p = 1 - q
p = = .9
H = 2pq = 2 x .9 x .1 = 0.18
Nheterozygous = .18 x 508 = 91
29ez - 2

22. P = allelic diversity = percent polymorphism
H and P
H = heterozygosity = the percent of heterozygous genotypes in
the population for that locus
H = 2pq (for 2 allele case)
H = 2pq + 2qr + 2 pr (for 3 allele case)
P = allelic diversity = percent polymorphism
= percent of loci for which alternative alleles exist in
the population
29A

23. Gene pools Population 1 N = 8 (7 homozygous) 2N = 16 alleles
f (blue) = 1/16
f (red) = 15/16
Population 2
N = 8
f (blue) = 16/16
Polymorphic locus
typical
Monomorphic locus
population P = approximately 0.25
individual H = approximately 0.07 5f

24. The relationship of P to H
Possible alleles = a, b, c aa bb ab ab cc ac ac ac
H = 1/4 H = 4/4
N alleles = 3 N alleles = 3
Conclusion: H is not sensitive to the number of different alleles
for the locus.
30f

25. The relationship of H to P
Population A, locus X
alleles frequency a b
H = 2pq = .50
P is low
Population B, locus X
alleles frequency
a .7
b .05
c .05
d .05
e .05
f .05
g .05
H = .495, P is high
30A

26. Uses of molecular genetics* in conservation
1) Parentage and kinship
2) Within-population genetic variability
3) Population structure and intraspecific phylogeny
4) Species boundaries, hybridization phenomena,
and forensics
5) Species’ phylogenies and macroevolution
*e.g., electrophoresis
protein sequencing
DNA fingerprinting
immunological techniques
15 -2

27. From Encarta

28. Western Pyrénées National Park, France
From Encarta

29. Effective population size
N = population size = total number of individuals
Ne = effective population size
= ideal population size that would have a rate of decrease in H equal to that of the actual population (N)
number of individuals contributing gametes to the next generation
32A-1

30. Effective population size
Predictable loss of heterozygosity (H) in each generation for non-ideal populations
GLT Ne = .32 N; N = 100, Ne = 32
Loss of H(N) = loss of H (Ne)
If Ne/N 1, then rate of loss of H is minimum.
The larger the Ne, the lower the rate of loss of H. 1
Rate of loss of H defined: 2Ne per generation
32b

31. Examples of effective population size
Taxon Ne
Drosophila .48 to .71 N
Humans to .95 N
a snail species .75 N
plants lower
golden lion tamarins .32 N (94 of 290)
32A-2

32. Assumptions of an ideal population
Infinitely large population
random mating
no mutation
no selection
no migration
31a -1

33. 5 causes of microevolution
1) genetic drift - stochastic variation in inheritance
Expected F2:
Observed F2:
2) Assortative (nonrandom) mating
3) Mutation
4) Natural selection
5) Migration (gene flow)
Random deviation
31a-2

34. F1 allele frequencies = Parental allele frequencies
Sampling Error
F1 allele frequencies = Parental allele frequencies
Caused by, for example:
Behavioral traits producing assortative mating
Genetic stochasticity
Results in Genetic Drift
= random deviation from expected allele frequencies
34A-2

35. Fixation of alleles Fn fixed Parental generation for many populations
lost
A = .5
a = .5
Genetic
drift
p = 1.0
A = 0
a = 1.0
lost
p = q = .5
fragmentation
fixed
q = 1.0
time
34A-1

36. What is a formula for calculating the effect of unequal
numbers of males and females (non-random breeding)on Ne?
Ne = MF M = # of breeding males
M + F F = # of breeding females
Population A Population B
M = M = 10
F = F = 90
N = 100
Ne = 4 x 50 x = 4 x 10 x 90
= = 36
10f

37. The effect of non-random mating on H
Given 2 cases, with N = 150 and Ne = 100 (population A)
Ne = (population B)
Ht=1 = Ht = the proportion of heterozygosity
2 Ne remaining in the next (t=1) generation
Population A: Ht = = = .995
2 x 100
Population B: Ht = = = .986
2 x 36 *
% H
remaining
after
t=1
generations
36A-1

38. * Generalized equation: Ht = H0 1 - 1 t t = # of generations later
2Ne H0 = original heterozygosity
What is H after 5 more generations?
Population A: H5 = H0 (.995) 5 = .995 (.995)5 = .970
Population B: H5 = H0 (.986) 5 = .995 (.986)5 = .919 *
36A-2

39. Formulae for calculating H and Ne
1 = proportion of H0 lost at each generation
2Ne
= proportion of H0 remaining after the first generation
2 Ne
Ht = H t = the absolute amount of H0 remaining after
2Ne t generations
Ne = 4 MF 1) unequal sex ratios or
M + F 2) nonrandom breeding
decrease Ne
37A

40. Mutation Nondisjunctive point mutations
over short term: not important in changing allele frequencies
f (A1) = mutation rate A1 --> A2 =
105
over 2000 generations, f (A1) = 0.49
If f (A2) increases rapidly, selection must be involved
Long-term, over evolutionary time
mutation is critical - providing raw material for natural selection
Mutation rate is independent of H, P, Ne
but mutation can increase H and increase P
36A1