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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 SS SF FF TotalNumber 36 47 23 106Frequence 0.34 0.44 0.22 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 SS SF FF TotalAntal 36 47 23 106Frekvens 0,34 0,44 0,22 1,00
Calculation of gene frequencies
• Gene frequency derived from numbers
• Gene frequency derived from proportions
Gene frequency derived from numbers
• S: p = (236+47)/(2106) = 0.56
• F: q = (223+47)/(2106) = 0.44» Total:
p + q = 1.00
Genotype SS SF FF TotalNumber 36 47 23 106Frequency 0.34 0.44 0.22 1.00
Gene frequency derived from proportions
• S: p = 0.34+0.50.44 = 0.56
• F: q = 0.22+0.50.44 = 0.44– Total: p + q = 1.00
SS SF FF Total36 47 23 106
0.34 0.44 0.22 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 = (22+18)/(243) = 0.256
• Allele frequency of ”199”: q = (20+12)/(243) = 0.140
• Allele frequency of ”195”: r = 1 - p - q = 0.604
Dominant inheritance
Phenotype Sort Gul TotalGenotype EE+Ee eeNumber 182 18 200Frequency 0.91 0.09 1.00
Gene frequency calculation for dominant inheritance
• q 2 = qq = 18/200 = 0.09
• q = qq = 0.30
• p = 1-q = 1-0.30 = 0.70
Phenotype Sort Gul TotalGenotype EE+Ee eeNumber 182 18 200Frequence 0.91 0.09 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
• SS: pp = 0.560.56 = 0.314
• FF: qq = 0.440.44 = 0.194
• SF: 2pq = 2 0.560.44 = 0.493
Genotypefrekvens:
Genotype SS SF FF TotalNumber obs. 36 47 23 106 =NFrequency exp. p*p 2pq q*q 1,00Number exp. 33.2 52.3 20.5 106
2-test for H-W equilibrium• H0: No difference between observed and expected
numbers 2 = (O-E)2/E = 1.09
• Significant level: = 0.05
• Degrees of freedom: df = 1
Genotype SS SF FF TotalNumber, obs. 36 47 23 106 =NFrekvens exp. 0.314 0.493 0.194 1.00Number, exp. 33.2 52.3 20.5 106O-E, deviation 2.8 -5,4 2.5 ~
(O-E)2/E 0.24 0.54 0.31 1.09
2-test
• P > 0.20 P > • 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 inheritanceX-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 = (23+53)/(2173) = 0.17
• ofemale: q = (2117+53)/(2173) = 0.83
• Omale: p = 28/177 = 0.16
• omale: q = 149/177 = 0.84
Sex Females Malesr Genotype OO Oo oo Total O o TotalNumber 3 53 117 173 28 149 177Frequency 0.02 0.31 0.67 1.00 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 Frequency AA AA p2 p2 = p4
AA Aa 2 p2 2pq = 4p3 qAA aa 2 p2 q2 = 2p2 q2
Aa Aa 2pq 2pq = 4p2 q2
Aa aa 2 2pq q2 = 4pq3
aa aa q2 q2 = 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
gene A/gene B B b FrequencyA r = p(A) p(B) + D s = p(A) q(b) - D p(A)a t = q(a) p(B) - D u = q(a) q(b) + D q(a)
Frequency p(B) q(b) 1
Linkage Rekombination Repulsion
Genotype process Genotype
A B A B A b
a b a b a B
Gamete frequencies by linkage fits into a two by two table
Gamete frequencies by linkage: Calculation example
• Test for independence
• H0: D = 0, 2 = 9.7, df = 1, = 0.05• H0 rejected linkage disequilibrium• D = r - p(A) p(B)
= 0.21-0.7 0.4 = - 0.07
gene A/gene B B b Sum (Freq)A 21 (r=0.21) 49 (s=0.49) 70 p(A)=0.7a 19 (t=0.19) 11 (u=0.11) 30 q(a)=0.30
Sum (Freq) 40 p(B)=0.4 60 q(b)=0.6 100 1
Gamete frequencies by linkage
• The gametes Ab og aB are in repulsions phase• Obs. - Exp. = deviation = D
Gamet Obs. Frequency Exp. Frequency DeviationAB r p(A) p(B) DAb s p(A) q(b) - DaB t q(a) p(B) - Dab u q(a) q(b) D
Linkage Rekombination Repulsion
Genotype process Genotype
A B A b
a b a B
a b a b a B
Gamete frequencies by linkage
Gamet AB / r Ab / s aB / t ab / uAB / r AABB / rr AABb / sr AaBB / rt AaBb / ruAb / s AABb / sr Aabb / ss AaBb / st Aabb / suaB / t AaBB / tr AaBb / ts aaBB / tt aaBb / tuab / u AaBb / ur Aabb / us aaBb / ut aabb / uu
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