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Chapter 2: Hardy-Weinberg
• Gene frequency
• Genotype frequency
• Gene counting method
• Square root method
• Hardy-Weinberg low
• Sex-linked inheritance
• Linkage and gamete frequency
Co-dominant inheritance
• S is the ”slow”
albumin allele
• F is the ”fast”
albumin allele
Genotyper
Genotype frequency
Genotype
Number
Frequence
SS
36
0.34
SF
47
0.44
FF
23
0.22
Total
106
1.00
Calculation of genotype frequencies
• Genotype frequency of SS: 36/106 = 0.34
• Genotype frequency of SF: 47/106 = 0.44
• Genotype frequency of FF: 23/106 = 0.22
Genotype
Antal
Frekvens
SS
36
0,34
SF
47
0,44
FF
23
0,22
Total
106
1,00
Calculation of gene frequencies
• Gene frequency derived from numbers
• Gene frequency derived from proportions
Gene frequency derived from
numbers
• S:
• F:
p = (236+47)/(2106) = 0.56
q = (223+47)/(2106) = 0.44
» Total:
Genotype
Number
Frequency
SS
36
0.34
p + q = 1.00
SF
47
0.44
FF
23
0.22
Total
106
1.00
Gene frequency derived from
proportions
• S:
• F:
p = 0.34+0.50.44 = 0.56
q = 0.22+0.50.44 = 0.44
– Total:
SS
36
0.34
p + q = 1.00
SF
47
0.44
FF
23
0.22
Total
106
1.00
Multiple alleles
• The calculation of gene frequency for more than
two alleles
Gene frequency calculation for
multiple alleles
• Allele frequency of ”209”:
p = (22+18)/(243) = 0.256
• Allele frequency of ”199”:
q = (20+12)/(243) = 0.140
• Allele frequency of ”195”:
r = 1 - p - q = 0.604
Dominant inheritance
Phenotype
Genotype
Number
Frequency
Sort
EE+Ee
182
0.91
Gul
ee
18
0.09
Total
200
1.00
Gene frequency calculation for
dominant inheritance
• q 2 = qq = 18/200 = 0.09
• q = qq = 0.30
• p = 1-q = 1-0.30 = 0.70
Phenotype
Genotype
Number
Frequence
Sort
EE+Ee
182
0.91
Gul
ee
18
0.09
Total
200
1.00
Hardy-Weinberg law
• The frequency of homozygotes is equal to
the gene frequencies squared: p2 og q2
• The frequency of heterozygotes is equal to
twice the product of the two gene
frequencies: 2pq
• Gene- and genotype frequencies are
constant from one generation to the next
Hardy-Weinberg law
Genotypefrekvens:
• SS: pp = 0.560.56 = 0.314
• FF: qq = 0.440.44 = 0.194
• SF: 2pq = 2  0.560.44 = 0.493
Genotype
Number obs.
Frequency exp.
Number exp.
SS
36
p*p
33.2
SF
47
2pq
52.3
FF
23
q*q
20.5
Total
106 =N
1,00
106
2-test for H-W equilibrium
• H0: No difference between observed and expected
numbers
• 2 = S (O-E)2/E = 1.09
• Significant level: a = 0.05
• Degrees of freedom: df = 1
Genotype
Number, obs.
Frekvens exp.
Number, exp.
O-E, deviation
(O-E)2/E
SS
36
0.314
33.2
2.8
0.24
SF
47
0.493
52.3
-5,4
0.54
FF
23
0.194
20.5
2.5
0.31
Total
106 =N
1.00
106
~
1.09
2-test
• P > 0.20
P>a
• H0 is not rejected. There is no significant
difference between observed and expected
numbers
Conclusion: There is no significant deviation
from Hardy-Weinberg equlibrium for
albumin type in Danish German Shepherd
dogs
Sex-linked inheritance
X-linkage
• Males an females do not necessarily contain the
same gene frequencies
• The mammalian male’s X chromosome comes
from the mother
• In the mammalian male expression of the gene is
direct, i.e. the genotype frequency is equal to the
gene frequency
• The genotype in the male is called a hemi zygote
The Orange gene in cats
• XX-individuals:
OO gives orange coat colour
Oo gives mixed coat colour
oo gives no orange colour in
the coat
• XY-individuals:
• O gives orange coat colour
• o gives no orange colour in the
coat
Calculation of the frequency of
the orange gene in cats
• Ofemale: p = (23+53)/(2173) = 0.17
• ofemale: q = (2117+53)/(2173) = 0.83
• Omale: p = 28/177 = 0.16
• omale: q = 149/177 = 0.84
Sex
Genotype
Number
Frequency
OO
3
0.02
Females
Oo
oo
53
117
0.31 0.67
Total
173
1.00
Malesr
O
o
Total
28
149
177
0.16 0.84 1.00
Sex-linked inheritance
• Sex-linked recessive diseases can be
expected to occur at a higher frequency in
males compared to the females
• Males: Gene frequency q = 0.01
Genotype frequency = Gene frequency
• Females: Gene frequency q = 0.01
Genotype frequency = q2 = 0.0001
Mating type frequencies at
random mating
Mating type
AA  AA
AA  Aa
AA  aa
Aa  Aa
Aa  aa
aa  aa
Frequency
p2  p2
2  p2  2pq
2  p2  q2
2pq  2pq
2 2pq q2
q2  q2
= p4
= 4p3  q
= 2p2  q2
= 4p2  q2
= 4pq3
= q4
Mating type frequencies
• Mono genetic
inherited diseases
• Closely related dog
breeds:
gene frequencies
Gamete frequencies, linkage
and linkage disequilibrium
• Gamete frequencies are used when two
genes at two loci are studied simultaneously
• A marker allele always occurs with a
harmful gene on the other locus
Gamete frequencies by linkage
fits into a two by two table
Linkage
Rekombination
Repulsion
Genotype
process
Genotype
A
B
A
B
A
b
a
b
a
b
a
B
gene A/gene B
A
a
Frequency
B
r = p(A) p(B) + D
t = q(a) p(B) - D
p(B)
b
s = p(A) q(b) - D
u = q(a) q(b) + D
q(b)
Frequency
p(A)
q(a)
1
Gamete frequencies by linkage:
Calculation example
• Test for independence
gene A/gene B
A
a
Sum (Freq)
B
21 (r=0.21)
19 (t=0.19)
40 p(B)=0.4
b
49 (s=0.49)
11 (u=0.11)
60 q(b)=0.6
• H0: D = 0, 2 = 9.7, df = 1, a = 0.05
• H0 rejected  linkage disequilibrium
• D = r - p(A)  p(B)
= 0.21-0.7  0.4 = - 0.07
Sum (Freq)
70 p(A)=0.7
30 q(a)=0.30
100 1
Gamete frequencies by linkage
Gamet
AB
Ab
aB
ab
Obs. Frequency
r
s
t
u
Exp. Frequency
p(A) p(B)
p(A) q(b)
q(a) p(B)
q(a) q(b)
Deviation
D
-D
-D
D
• The gametes Ab og aB are in repulsions phase
• Obs. - Exp. = deviation = D
Gamete frequencies by linkage
Linkage
Rekombination
Genotype
Repulsion
process
Genotype
A
B
A
b
a
b
a
B
Gamet
AB / r
Ab / s
aB / t
ab / u
AB / r
AABB / rr
AABb / sr
AaBB / tr
AaBb / ur
Ab / s
AABb / sr
Aabb / ss
AaBb / ts
Aabb / us
aB / t
AaBB / rt
AaBb / st
aaBB / tt
aaBb / ut
ab / u
AaBb / ru
Aabb / su
aaBb / tu
aabb / uu
Linkage disequilibrium
• Obs - Exp = deviation = D
• D = u - q(a)q(b), or
D = ru - ts (= (f (AB/ab) - f (Ab/aB))/2 )
• Maximum disequilibrium (Dmax) occurs
when all double heterozygotes are either in
linkage phase (AB/ab) or in repulsions
phase (Ab/aB). Dmax = 0.5
Disappearance of linkage
disequilibrium
• Dn = D0(1-c)n, where D0 is the linkage
disequilibrium in the base population
Gamete frequencies at linkage
disequilibrium and equilibrium
• In connection with a new mutation, linkage
disequilibrium occurs in many of the
following generations, as the mutation only
arises in one chromosome
• There is always maximum linkage
disequilibrium within a family