Hardy-Weinberg Equilibrium from the random pairing of gametes
A
a
Aa
Aa
f(A) = p = 120/200 = .60
f(a) =q = 80/200 = .40
Gene
pool
‘broadcast spawners’
AA
A AAaa
a AA
A
A a a a
a a AAA
f(A)=.6
Assumptions:
Diploid, sexual
f(a)=.4
Random mating
Infinite population size
No: selection, mutation migration
f(A)=.6
f(a)=.4
.6x.6=p2
.36
.6x.4=pq
.24
.6x.4=pq
.24
.4x.4=q2
.16
p2 + 2pq + q2 = 1
Hardy - Weinberg Law
Genotype frequencies are predicted from products of allele
frequencies
Genotype frequencies stay constant under random mating
One generation of random mating restores HWE
Homework: prove that one
generation of random
mating generated HWE:
f(AA) = .3
f(Aa) = .2
f(aa) = .5 give allele freqs.
p = 0.55, q = 0.45
Godfray Harold Hardy
Wilhelm Weinberg
f(AA) = .4
f(Aa) = .3
f(aa) = .3 give allele freqs.
p = 0.55, q = 0.45
http://images.google.com/images?svnum=10&hl=en&lr=&ie=ISO-8859-1&q=Godfrey+Harold+Hardy
Hardy Weinberg proportions from the random pairing of diploid individuals
f(A1) = p = P + 1/2H
f(A2) = q = Q + 1/2H
Maximum heterozygosity occurs at p = q = 0.5
Heterozygosity with multiple alleles
n
H E  1  p 2i
i 1
Maximum heterozygosity occurs when all alleles are equally frequent
With n alleles, each allele has a frequency of 1/n

1 2
HE  1  
n 
i1
n
1 2
H E  1 n 
n 
(n 1)
HE 
n
3 alleles: Hmax = 0.667; 4 alleles: Hmax = 0.75



X-linked or Haplo-Diploid Genotype Frequencies
X
X
X
Y
Female genotypes
Pf = f(A1A1)
Hf = f(A1A2)
Qf = f(A2A2)
Male genotypes
Pm = f(A1)
Qm = f(A2)
Female A2 alleles
qf = Qf +1/2Hf
Male A2 alleles
qm = Qm
Average allele frequency
2
1
q  q f  qm
3
3

Allele frequency in the next generation
q’ = average of the male and female freqs.
qf 
1
q f  qm 

2
Female gets one X
From each parent

q
m  qf

Male gets X
From mom
X-linked allele frequencies in males and females
Starting from alternative alleles in each sex
X1
X1
pm’ : Male gets X from mom
pf’ : Female gets 1 X from mom, 1 from dad
X2
Y
Nucleotide variation at the Adh locus in Drosophila
2659 pase pairs: 52 variable sites in 11 alleles
Adh locus
Fast/Slow
polymorphism
Each position in the sequence is an allele, mostly 2 alleles per position
Each sequence is a haplotype = a haploid type, many haplotypes per locus
Nucleotide polymorphism
pˆ s 
S
N
= 52/2659 = 0.01956

Nucleotide diversity = sum of nucleotide differences among all pairs
number of pairwise comparisons
   pi p j  ij
ij
= 0.0065 ± 0.0017
Autosomes
Sex chromosomes
Female
A
a
X
X
Male
A
a
X
Y
mtDNA
VNTR Allele Frequency Distributions in Different Human Populations
0.16
Arizona Caucasians D1S7 Hae III
0.20
n = 270 H = 0.94
Frequency
Frequency
0.20
0.12
0.08
0.04
0.00
0.04
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31
Fixed Bin
Swiss Caucasians D1S7Hae III
0.20
n = 410 H = 0.94
0.08
0.04
0.00
0.16
Swiss Caucasians D2S44Hae III
n = 804 H = 0.93
0.12
0.08
0.04
0.00
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31
Fixed Bin
Fixed Bin
Orange Co. Chinese D1S7 Hae III
0.20
n = 218 H = 0.94
Frequency
Frequency
0.08
Fixed Bin
0.12
0.16
0.12
0.00
0.16
0.20
n = 270 H = 0.92
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31
Frequency
Frequency
0.20
0.16
Arizona Caucasians D2S44 Hae III
0.12
0.08
0.04
0.00
Orange Co. Chinese D2S44Hae III
n = 216 H = 0.88
0.16
0.12
0.08
0.04
0.00
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31
Fixed Bin
Fixed Bin
HWE is used to calculate the probability of a ‘match’ for DNA profiles
With multiple, truly independent loci, individuals can be uniquely identified
Population
Allele
Arizona Caucasian
OC Chinese
Pooled
D1S7 locus
7
26 H=2pq
0.007 0.115 0.00161
0.018 0.041 0.00148
0.013 0.078 0.00195
D2S44
10
0.104
0.213
0.159
locus
18 H=2pq
0.081 0.01685
0.005 0.00213
0.043 0.01363
Prob. of 1 chance
match
in:
2.71E-05
36,866
3.14E-06
318,078
2.66E-05
37,622