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

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

Calculation of gene frequencies Gene frequency derived from numbers Gene frequency derived from proportions

Gene frequency derived from numbers S: p = (2  36+47)/(2  106) = 0.56 F: q = (2  23+47)/(2  106) = 0.44 » Total: p + q = 1.00

Gene frequency derived from proportions S: p =  0.44 = 0.56 F: q =  0.44 = 0.44 –Total: p + q = 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) = Allele frequency of ”199”: q = (2  0+12)/(2  43) = Allele frequency of ”195”: r = 1 - p - q = 0.604

Dominant inheritance

Gene frequency calculation for dominant inheritance q 2 = q  q = 18/200 = 0.09 q =  q  q = 0.30 p = 1-q = = 0.70

Hardy-Weinberg law The frequency of homozygotes is equal to the gene frequencies squared: p 2 og q 2 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: p  p = 0.56  0.56 = FF: q  q = 0.44  0.44 = SF: 2pq = 2  0.56  0.44 = Genotypefrekvens:

 2 -test for H-W equilibrium H 0 : No difference between observed and expected numbers  2 =  (O-E) 2 /E = 1.09 Significant level:  = 0.05 Degrees of freedom: df = 1

 2 -test P > 0.20 P >  H 0 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 O female : p = (2  3+53)/(2  173) = 0.17 o female : q = (2  )/(2  173) = 0.83 O male : p = 28/177 = 0.16 o male : q = 149/177 = 0.84

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 = q 2 =

Mating type frequencies at random mating Mating type Frequency AA  AA p 2  p 2 = p 4 AA  Aa2  p 2  2pq= 4p 3  q AA  aa2  p 2  q 2 = 2p 2  q 2 Aa  Aa 2pq  2pq= 4p 2  q 2 Aa  aa2  2pq  q 2 = 4pq 3 aa  aa q 2  q 2 = q 4

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

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 H 0 : D = 0,  2 = 9.7, df = 1,  = 0.05 H 0 rejected  linkage disequilibrium D = r - p(A)  p(B) =  0.4 =

Gamete frequencies by linkage The gametes Ab og aB are in repulsions phase Obs. - Exp. = deviation = D

Linkage Rekombination Repulsion Genotype process Genotype A B A b a b a B a b a b a B a b a b a B Gamete frequencies by linkage

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 (D max ) occurs when all double heterozygotes are either in linkage phase (AB/ab) or in repulsions phase (Ab/aB). D max = 0.5

Disappearance of linkage disequilibrium D n = D 0 (1-c) n, where D 0 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