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