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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.

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chapter 2 hardy weinberg
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
Co-dominant inheritance
  • S is the ”slow” albumin allele
  • F is the ”fast” albumin allele

Genotyper

calculation of genotype frequencies
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
Calculation of gene frequencies
  • Gene frequency derived from numbers
  • Gene frequency derived from proportions
gene frequency derived from numbers
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
Gene frequency derived from proportions
  • S: p = 0.34+0.50.44 = 0.56
  • F: q = 0.22+0.50.44 = 0.44
    • Total: p + q = 1.00
multiple alleles
Multiple alleles
  • The calculation of gene frequency for more than two alleles
gene frequency calculation for multiple 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
gene frequency calculation for 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
Hardy-Weinberg law
  • The frequency of homozygotes is equal to the gene frequenciessquared: 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 law13
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
2 test for h w equilibrium
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
2 test
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 albumintype inDanish German Shepherd dogs

sex linked inheritance x linkage
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 hemizygote
the orange gene in cats
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
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 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
slide20

Mating type frequencies at random matingMating type FrequencyAA  AA p2 p2 = p4AA  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

Mating type frequencies
gamete frequencies linkage and linkage disequilibrium
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
Gamete frequencies by linkage fits into a two by two table

Linkage Rekombination Repulsion

Genotypeprocess Genotype

A B A B A b

a b a b a B

gamete frequencies by linkage calculation example
Gamete frequencies by linkage: Calculation example
  • Test for independence
  • H0: D = 0, c2 = 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
gamete frequencies by linkage
Gamete frequencies by linkage
  • The gametes Ab og aB are in repulsions phase
  • Obs. - Exp. = deviation = D
gamete frequencies by linkage26
Gamete frequencies by linkage

Linkage Rekombination Repulsion

Genotypeprocess Genotype

A BA b

ab aB

a b a b a B

linkage disequilibrium
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
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
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