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
p2 + 2pq + q2 = 1
f [AA] + 2 [f(Aa)] + f [aa] = 1
A
a
A
AA Aa
a
Aa
aa
1
26a
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
2
27a-1
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!!
3
27a-2
Hardy-Weinberg Equilibrium
Conclusions:
1)Allele frequencies are conserved (i.e., the
same) from one generation to the next.
2) genotype frequency reaches HardyWeinberg equilibrium in one generation
4
27a-2
Hardy-Weinberg’s caveats:
But
Hardy-Weinberg Law: allele frequencies in a population
remain constant from generation to generation …..
Only
•IF
•IF
•IF
random mating
all genotypes are equally viable
not disturbed by mutation, selection or whatever
Bottom line: Only in an IDEAL population is genetic diversity conserved forever.
5
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
R r r R RR R r r Rr
10 genotypes
r R R R r R r R R
20 alleles
Parental gametes
Probability of
F1
r = .4; R =.6
6
25A -1
F1 genotypes and phenotypes
Genotype Frequency
Rr or rR
.48
rr
.16
RR
.36
-------------
Phenotype Frequency
R_ .48+.36=.84
rr
.16
--------------=1
=1
7
25A -2
If allele frequencies are P and Q in the parental generation, how
do we calculate what they will be in the F1 generation?
8
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
9
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?
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 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
11
What will be the expected phenotypes and their ratios in this
example?
12
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= .49 + .42 = .91
aa
= .09
13
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
14
29f
H =?
Aa
aa
aa
AA
aa
aA
aa
AA
aA
AA
aa
aa
AA
aa
aA
H = 4/15
15
Heterozygosity defined
H (“H-bar”) = average heterozygosity for all loci in a
population.
H estimated = % heterozygous loci
those examined
H = 2pq
16
29f
Calculating H
(assuming simple dominance and Hardy-Weinberg Eq.)
calculate H if q2 = 0.09
q2 = 0.09
p2 = 0.49
2 pq = 0.42
H = 2pq = 0.42 (for only 2 alleles)
f (a) = 0.3 = q
f (A) = 0.7 = p
17
29a - 1
Calculating H
(but if……)
Codominance
Genotype
AA
Aa
aa
Phenotype N
Red
50
Pink
22
White
10
total = 82
H = 22/82 (don’t need Hardy - Weinberg)
18
29a - 1
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
p
q
p
pp
pq
pr
q
pq
qq
qr
r
pr
qr
rr
r
19
29a
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
F1 generation p2 = ?
2pq = ?
q2 = .01
Aa
aa
= f (AA)
= f (Aa) = H
= f (aa)
20
29ez - 1
q2 = .01
q = .01 = 0.1
p + q = 1 so p = 1 - q
p = 1 - .1 = .9
H = 2pq = 2 x .9 x .1 = 0.18
Nheterozygous = .18 x 508 = 91
21
29ez - 2
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
22
Gene pools
Population 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
typical
population P = approximately 0.25
individual H = approximately 0.07
Polymorphic locus
Monomorphic locus
23
The relationship of P to H
Possible alleles = a, b, c
aa
bb
ab
ab
cc
ac
ac
ac
H = 1/4
N alleles = 3
H = 4/4
N alleles = 3
Conclusion: H is not sensitive to the number of different alleles
for the locus.
24
30f
The relationship of H to P
Population A, locus X
alleles
a
b
frequency
.5
.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
25
30A
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
26
15 -2
27
Western Pyrénées National Park, France
28
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
29
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
30
32b
Examples of effective population size
Taxon
Drosophila
Humans
a snail species
plants
golden lion tamarins
Ne
.48 to .71 N
.69 to .95 N
.75 N
lower
.32 N (94 of 290)
31
32A-2
Assumptions of an ideal population
•
•
•
•
•
Infinitely large population
random mating
no mutation
no selection
no migration
32
31a -1
5 causes of microevolution
1) genetic drift - stochastic variation in inheritance
Expected F2: 9 - 3 - 3 - 1
Random deviation
Observed F2: 9 - 3 - 2.8 - 1.2
2) Assortative (nonrandom) mating
3) Mutation
4) Natural selection
5) Migration (gene flow)
33
31a-2
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
34
34A-2
Fixation of alleles
Fn
Parental generation
for many populations
A = .5
a = .5
p = q = .5
fixed
A = 1.0
a=0
Genetic
drift
lost
p = 1.0
A=0
a = 1.0
lost
fixed
q = 1.0
time
35
34A-1
What is a formula for calculating the effect of unequal
numbers of males and females (non-random breeding)on Ne?
Ne =
4 MF
M+F
M = # of breeding males
F = # of breeding females
Population A
M = 50
F = 50
Population B
M = 10
F = 90
N = 100
Ne = 4 x 50 x 50
50 + 50
= 100
= 4 x 10 x 90
10 + 90
= 36
36
10f
The effect of non-random mating on H
Given 2 cases, with N = 150 and
*
Ht=1 = 1 - 1
2 Ne
Ne = 100 (population A)
Ne = 36 (population B)
Ht = the proportion of heterozygosity
remaining in the next (t=1) generation
Population A: Ht = 1 - 1
2 x 100
Population B: Ht = 1 -
1
2 x 36
= 1 - .005 = .995
= 1 - .014 = .986
%H
remaining
after
t=1
generations
37
36A-1
Generalized equation:
*
Ht = H0 1 -
1
2Ne
t
t = # of generations later
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
38
36A-2
Formulae for calculating H and Ne
1 = proportion of H0 lost at each generation
2Ne
1 - 1 = proportion of H0 remaining after the first generation
2 Ne
Ht = H0 1 - 1
2Ne
Ne = 4 MF
M+F
t
= the absolute amount of H0 remaining after
t generations
1) unequal sex ratios or
2) nonrandom breeding
decrease Ne
39
37A
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 P
40
36A1