Chapter 2 HardyWeinberg Gene frequency Genotype frequency Gene

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: • F: p = (2 36+47)/(2 106) = 0. 56 q = (2 23+47)/(2 106) = 0. 44 » Total: p + q = 1. 00

Gene frequency derived from proportions • S: • F: – Total: p = 0. 34+0. 5 0. 44 = 0. 56 q = 0. 22+0. 5 0. 44 = 0. 44 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) = 0. 256 • Allele frequency of ” 199”: q = (2 0+12)/(2 43) = 0. 140 • Allele frequency of ” 195”: = 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 = 1 -0. 30 = 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: 2 pq • Gene- and genotype frequencies are constant from one generation to the next

Hardy-Weinberg law Genotypefrekvens: • SS: p p = 0. 56 = 0. 314 • FF: q q = 0. 44 = 0. 194 • SF: 2 pq = 2 0. 56 0. 44 = 0. 493

2 -test for H-W equilibrium • H 0: 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

2 -test • P > 0. 20 P>a • 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 • 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-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 = 0. 0001

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

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

Gamete frequencies by linkage: Calculation example • Test for independence • H 0: D = 0, 2 = 9. 7, df = 1, a = 0. 05 • H 0 rejected linkage disequilibrium • D = r - p(A) p(B) = 0. 21 -0. 7 0. 4 = - 0. 07

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

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

Linkage disequilibrium • Obs - Exp = deviation = D • D = u - q(a) q(b), or D = ru - ts (= (f (AB/ab) - f (Ab/a. B))/2 ) • Maximum disequilibrium (Dmax) occurs when all double heterozygotes are either in linkage phase (AB/ab) or in repulsions phase (Ab/a. B). Dmax = 0. 5

Disappearance of linkage disequilibrium • Dn = 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