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 = 1.0 (allele frequencies) p = 1 - q or, q = 1 - p

AA Aa

Aa aa

A

a

A a

p2 + 2pq + q2 = 1 f [AA] + 2 [f(Aa)] + f [aa] = 1

26a

2

Hardy-Weinberg Equilibrium

Parental generation: 2 alleles, r and R

f (R) + f (r) = 1.0

p + q = 1.0 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 = 0.01+.18/2= .1

q2 (.81) rr q = 0.81+.18/2 = .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

27a-2

Note: parental generation genotype frequencies do NOTpredict F1 generation genotype frequencies!!

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

Hardy-Weinberg Law: allele frequencies in a populationremain constant from generation to generation …..

•IF random mating•IF all genotypes are equally viable•IF not disturbed by mutation, selection or whatever

But

Only

Bottom line: Only in an IDEAL population is genetic diversity conserved forever.

Hardy-Weinberg’s caveats:

6

Sex = Random sampling of a gene poolPopulation of 10 individuals (N = 10)

Phenotypes Red WhiteWhite

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 rr R R R r R r R R

25A -1

7

F1 genotypes and phenotypes

Genotype Frequency Phenotype Frequency

Rr or rR .48 R_ .48+.36=.84

rr .16 rr .16

RR .36

25A -2

= 1

= 1-------------

---------------

8

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

9

If allele frequencies are P and Q in the parental generation, howdo 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 nextgeneration?

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 = pa = .3 = q

Expected genotypes and their frequencies in F1:AA = p2 = .49aa = q2 = .09Aa = 2pq = .42

12

What will be the expected phenotypes and their ratios in thisexample?

13

What will be the expected phenotypes and their ratios in thisexample?

Allele frequencies in P (parental generation):A = .7 = pa = .3 = q

Expected genotypes in F1:AA = p2 = .49aa = q2 = .09Aa = 2pq = .42

Expected phenotypes in F1:A‑ = .49 + .42 = .91aa = .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

Aa aa aa AA aa

aA aa AA aA AA

aa aa AA aa aA

H = 4/15

H = ?

16

Heterozygosity defined

H (“H-bar”) = average heterozygosity for all loci in a population.

H estimated = % heterozygous loci

those examined

29f

H = 2pq

17

Calculating H (assuming simple dominance and Hardy-Weinberg Eq.)

calculate H if q2 = 0.09

f (a) = 0.3 = q q2 = 0.09f (A) = 0.7 = p p2 = 0.49

2 pq = 0.42H = 2pq = 0.42 (for only 2 alleles)

29a - 1

18

Calculating H (but if……)Codominance Genotype Phenotype N

AA Red 50 Aa Pink 22 aa White 10

total = 82H = 22/82 (don’t need Hardy - Weinberg)

29a - 1

19

Calculating H for 3 alleles: p, q, r

p = 0.5

q = 0.4

r = 0.1

H = 2pq + 2qr + 2pr

= .40 + .08 + .10

= .58

pp pq pr

r

r

q

p

p q

pq qq qr

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 = .01 = f (aa)

29ez - 1

21

q = 01. = 0.1

p + q = 1 so p = 1 - qp = 1 - .1 = .9

H = 2pq = 2 x .9 x .1 = 0.18

Nheterozygous = .18 x 508 = 91

29ez - 2

q2 = .01

22

H and P

H = heterozygosity = the percent of heterozygous genotypes inthe 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 poolsPopulation 1

N = 8 (7 homozygous)

2N = 16 alleles

f (blue) = 1/16

f (red) = 15/16

Population 2

N = 8

2N = 16 alleles

f (blue) = 16/16

Polymorphic locus

Monomorphic locus

5f

population P = approximately 0.25

individual H = approximately 0.07

typical

24

The relationship of P to H

Possible alleles = a, b, c

Conclusion: H is not sensitive to the number of different alleles for the locus.

cc

aa

ac

ab

H = 1/4 H = 4/4N alleles = 3 N alleles = 3

bb

ac

ab

ac

30f

25

The relationship of H to P

Population A, locus X

alleles frequency

a .5

b .5

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

15 -2

*e.g., electrophoresisprotein sequencingDNA fingerprintingimmunological techniques

27

From Encarta

28Western 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 generation32b

31

Examples of effective population size

Taxon Ne

Drosophila .48 to .71 N

Humans .69 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: 9 - 3 - 3 - 1

Observed F2: 9 - 3 - 2.8 - 1.2

2) Assortative (nonrandom) mating

3) Mutation

4) Natural selection

5) Migration (gene flow)

31a-2

Random deviation

34

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

Parental generationfor many populations

A = .5a = .5

p = q = .5

frag

men

tati

on

Fn

A = 1.0a = 0

fixed

lost

fixed

A = 0a = 1.0

lost

q = 1.0

p = 1.0Genetic drift

34A-1

time

36

What is a formula for calculating the effect of unequal numbers of males and females (non-random breeding)on Ne?

Ne = 4 MF M = # of breeding malesM + F F = # of breeding females

Population A Population BM = 50 M = 10F = 50 F = 90

N = 100Ne = 4 x 50 x 50 = 4 x 10 x 90

50 + 50 10 + 90

= 100 = 3610f

37

The effect of non-random mating on H

Given 2 cases, with N = 150 and Ne = 100 (population A)Ne = 36 (population B)

Ht=1 = 1 - 1 Ht = the proportion of heterozygosity 2 Ne remaining in the next (t=1) generation

Population A: Ht = 1 - 1 = 1 - .005 = .995 2 x 100

Population B: Ht = 1 - 1 = 1 - .014 = .986 2 x 36

% Hremainingaftert=1generations

*

36A-1

38

36A-2

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

*

39

Formulae for calculating H and Ne

37A

1 = proportion of H0 lost at each generation2Ne

1 - 1 = proportion of H0 remaining after the first generation 2 Ne

Ht = H0 1 - 1 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

40

Mutation

Nondisjunctive point mutations over short term: not important in changing allele frequencies

f (A1) = 0.5 mutation rate A1 --> A2 = 1 105

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 P36A1