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Chapter 4

Chapter 4. Genetic foundations of quantitative traits. Summarize chpt 3 …. When testing for Hardy-Weinberg equilibrium Calculate observed allele frequency by counting alleles or genotypes Calculate expected genotype frequency under Hardy-Weinberg equilibrium

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Chapter 4

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  1. Chapter 4 Genetic foundations of quantitative traits

  2. Summarize chpt 3 … • When testing for Hardy-Weinberg equilibrium • Calculate observed allele frequency by counting alleles or genotypes • Calculate expected genotype frequency under Hardy-Weinberg equilibrium • Compare expected and observed genotype frequency when testing for Hardy-Weinberg equilibrium

  3. The “one-locus model” to illustrate: genotypic value Quantitative genetics of a single locus

  4. Genotypic value GBB = o + a Gbb = o - a GBb = o + d • Origin = o • Additive effect = a • Dominance effect = d

  5. Genotypic value • Origin is middle of 2 homozygotes • Difference between 2 homozygotes = 2a • Heterozygote is not necessarily middle of 2 homozygotes

  6. Genotypic value • d ≠ 0  locus show dominance • d = a or d = --a  complete dominance • d > a  overdominant • d = 0  additive

  7. 115 125 122 Genotypic value origin: o = (bb+BB)/2 = (115+125)/2 = 120 additive effect: a = (BB – bb)/2 = (125 - 115)/2 = 5 dominance effect: d = Bb – o = 122 – 120 = 2

  8. Population mean: simple method.. • Suppose p(B) = 0.3, q(b) = 0.7 and random mating • BB = 125, Bb = 122, bb = 115 secs • What is population average? • fBB = p2 = 0.09; fBb = 2pq = 0.42; fbb = q2 = 0.49 • = fBB*BB+ fBb*Bb + fbb*bb • = 0.09*125 + 0.42*122 + 0.49*115 = 118.84 sec

  9. Population mean: general formula • Suppose p(B) = 0.3, q(b) = 0.7 and random mating • = p2 * GBB + 2pq * GBb + q2 * Gbb • = p2 * (o+a) + 2pq * (o+d) + q2 * (o-a) • = o+(p-q)a + 2pqd • = 120 +(0.3 -- 0.7)*5 + 2*0.3*0.7*2 = 118.84 sec Important message!! “p” is always assigned to the allele with the positive effect: +a

  10. The “one-locus model” to illustrate: • Breeding value

  11. We want to decrease mile time in Quarter horses How do we select the individuals that will give the best offspring? >> Breeding Value << Next generation

  12. the expected phenotypic value of offspring of the individual, when it is mated to randomly selected partners, expressed as a deviation from the population mean times 2. Breeding value = Expected offspring performance Breeding Value

  13. Breeding value “A” P = G + E Additive trait: G =  + A > A = G -  A = Difference between genotypic value and population mean

  14. Breeding Value 115 122 125 • p(B) = 0.3, q(b) = 0.7,  = 118.84 sec. • Breeding value of BB-individual, ABB? • Offspring: 30% will be BB, 70% will be Bb • E(Poff) = 0.3x125 + 0.7x122 = 122.9 • ABB = 2 x (122.9 - 118.84) = 8.12 sec.

  15. Allele substitution effect α Allele substitution effect α=[a+(q-p)d] =5.8 ABB = 2qα = 8.12 ABb = (q-p)α = 2.32 Abb = -2pα = -3.48 Avg breeding value = 0.09*8.12 + 0.42*2.32 + 0.49*(3.48) = 0 5.8 5.8 Breeding value is only additive Genotypic value is not

  16. Expected performance of offspring? 115 122 125  = 118.84; ABB = 8.12 s.; ABb = 2.32 s.; Abb = 3.48 s. Offspring performance?? BB-stallion: E(Poff) = 118.84 + ½ x 8.12 = 122.9 sec per mile Bb-stallion: E(Poff) = 118.84 + ½ x 2.32 = 120.0 sec per mile bb-stallion: E(Poff) = 118.84 + ½ x –3.48 = 116.7 sec per mile

  17. Summary on the breeding value (A) 1. The breeding value refers to the expected offspring performance 2. The average breeding value of all animals in a population is zero 3. The breeding value of Bb is exactly in the middle of BB and bb! 4. Genetic improvement  focus on breeding value

  18. The “one-locus model” to illustrate: • Dominance Deviation

  19. Dominance Deviation Additive trait: G =  + A Dominance (d  0): G =  + A + D D = “dominance deviation” D = Difference between genotypic value and breeding value

  20. No dominance deviation: d =0 α= a-(p-q)d = a! Breeding value ABB: 2qα = 2qa! Breeding value ABb: (q-p)a Breeding value Abb: -2pa GBB = u + A → G = o + a(p-q) + 2qa = o + a !

  21. Example: dominance deviation: d ≠ 0 G= µ + A + D Breeding value ABB = 2qα GBB = µ + ABB + DBB → G = [o + a(p-q) +2pqd] + 2qα + DBB (o + a) = [o + a(p-q) + 2pqd] + 2q[a-(p-q)d] + DBB => DBB = -2q²d

  22. DBB = -2q²d DBb = 2pqd Dbb = -2p²d D is the difference between G and A D is not inherited by offspring and is useless in breeding! Also homozygotes have D deviations! Dominance deviation

  23. p = 0.3 and d = 2 DBB = -2q2d = -2*0.72*2 = --1.96 Check: G = µ + A + D > 118.84 + 8.12 – 1.96 = 125! Derive dominance deviation for BB- quarterhorse 115 122 125

  24. Lets have a Break

  25. Exercise SCID in Arabian horses Allusive Gold (caspar): SCID free

  26. In 1978: estimated 28 % carriers In 1998: 15.1% (3926 horses) carriers, based on DNA test In 1998: 74 fowls died of SCID (no DNA test) Problem Do we need to get rid of all carriers? (proposed by DNA company)? Reduced the population size > increase risk of inbreeding How to get rid of SCID?

  27. Calculate observed frequency of SCID allele in the population, using 1998 data. Is the population in H-W equilibrium? Estimate the number of carrier horses in 1978, using data from 1998, and assuming 4 generations of 5 yrs. Questions

  28. Estimate the number of carrier horses in 2008, assume that we only use tested stallions to mate with randomly chosen mares (carrier and clear) Is this procedure effective? Do we need to start testing mares as well? Questions

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