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Assignment 7.5 Falconer & Mackay, chapter 7

Assignment 7.5 Falconer & Mackay, chapter 7. Sanja Franic VU University Amsterdam 2012. 2 alleles, A and B 3 situations: q=.2 q=.5 q=.8 mean phenotypic value per genotype: Question: What are the average effects of the two alleles, and the average effect of gene substitution? . AA.

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Assignment 7.5 Falconer & Mackay, chapter 7

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  1. Assignment 7.5Falconer & Mackay, chapter 7 Sanja Franic VU University Amsterdam 2012

  2. 2 alleles, A and B • 3 situations: • q=.2 • q=.5 • q=.8 • mean phenotypic value per genotype: • Question: What are the average effects of the two alleles, and the average effect of gene substitution? AA AB BB Genotype Genotypic value 122 154 188 Genotype frequency q2 2pq p2

  3. 2 alleles, A and B • 3 situations: • q=.2 • q=.5 • q=.8 • mean phenotypic value per genotype: • Question: What are the average effects of the two alleles, and the average effect of gene substitution? • Answer: AA AB BB Genotype Genotypic value 122 154 188 Genotype frequency q2 2pq p2 • = a + d (q - p) • aA = -pa • aB = qa • a=33 • d=-1

  4. 2 alleles, A and B • 3 situations: • q=.2 • q=.5 • q=.8 • mean phenotypic value per genotype: • Question: What are the average effects of the two alleles, and the average effect of gene substitution? • Answer: • Situation 1) • = a + d (q - p) = 33.6 • aA = -pa = -26.88 • aB= qa = 6.72 AA AA AB AB BB BB Genotype Genotype Phen. value Genotypic value 122 122 154 154 188 188 Genotype frequency q2 .32 2pq p2 Genotype freq .04 .64 • = a + d (q - p) • aA = -pa • aB = qa • a=33 • d=-1

  5. 2 alleles, A and B • 3 situations: • q=.2 • q=.5 • q=.8 • mean phenotypic value per genotype: • Question: What are the average effects of the two alleles, and the average effect of gene substitution? • Answer: • Situation 1) • = a + d (q - p) = 33.6 • aA = -pa = -26.88 • aB= qa = 6.72 • Situation 2) • = a + d (q - p) = 33 • aA = -pa = -16.5 • aB= qa = 16.5 AA AA AA AB AB AB BB BB BB Genotype Genotype Genotype Phen. value Phen. value Genotypic value 122 122 122 154 154 154 188 188 188 Genotype frequency q2 .5 .32 2pq p2 Genotype freq Genotype freq .04 .25 .64 .25 • = a + d (q - p) • aA = -pa • aB = qa • a=33 • d=-1

  6. 2 alleles, A and B • 3 situations: • q=.2 • q=.5 • q=.8 • mean phenotypic value per genotype: • Question: What are the average effects of the two alleles, and the average effect of gene substitution? • Answer: • Situation 1) • = a + d (q - p) = 33.6 • aA = -pa = -26.88 • aB= qa = 6.72 • Situation 2) • = a + d (q - p) = 33 • aA = -pa = -16.5 • aB= qa = 16.5 • Situation 3) • = a + d (q - p) = 32.4 • aA = -pa = -6.48 • aB= qa = 25.92 AA AA AA AA AB AB AB AB BB BB BB BB Genotype Genotype Genotype Genotype Phen. value Phen. value Phen. value Genotypic value 122 122 122 122 154 154 154 154 188 188 188 188 Genotype frequency q2 .5 .32 .32 2pq p2 Genotype freq Genotype freq Genotype freq .64 .04 .25 .25 .04 .64 • = a + d (q - p) • aA = -pa • aB = qa • a=33 • d=-1

  7. aA: mean deviation from the population mean of individuals who inherited allele A from one parent, the other allele having come at random from the population • so aA is a deviation from the population mean. What is the population mean? • m = a(p – q) + 2dpq • in our example (situation 1): • m = a(p – q) + 2dpq • = 33(.8-.2) + 2*(-1)*.8*.2 • = 19.48 • (on a scale where 0 is midpoint!) • our scale: • m = 19.48 + midpoint • = 19.48 + 155 • = 174.48 A (q) AA (q) q*(-a) A -qa + pd aA= -qa + pd - m aA = -p[a+ d(q - p)] AA AB BB Genotype B (p) AB (p) p*d Genotypic value 122 154 188 Genotype frequency q2 2pq p2 aA aB • aA = -pa = -26.88 -> 174.48-26.88 = 147.68 • aB= qa = 6.72 -> 174.48+6.72 = 181.2 m

  8. Situation 1) • aA = -pa = -26.88 -> 174.48-26.88 = 147.68 • aB= qa = 6.72 -> 174.48+6.72 = 181.2 • f(B)=.8 • f(A)=.2 • Situation 2) • aA = -pa = -16.5 -> 154.5-16.5 = 138 • aB= qa = 16.5 -> 154.5+16.5 = 171 • f(B)=.5 • f(A)=.5 • Situation 3) • aA = -pa = -6.48 -> 134.88-6.48 = 128.4 • aB= qa = 25.92 -> 134.88+25.92 = 160.8 • f(B)=.2 • f(A)=.8 AA AA AA AB AB AB BB BB BB Genotype Genotype Genotype Genotypic value Genotypic value Genotypic value 122 122 122 154 154 188 188 188 Genotype frequency Genotype frequency Genotype frequency q2 q2 q2 2pq 2pq 2pq p2 p2 p2 aA aA aB aA aB aB m m m

  9. A simplified example involving height

  10. A simplified example involving height, d=0:

  11. A simplified example involving height, d=0: AA AB BB Genotype Genotypic value 160 180 200 Genotype frequency q2 2pq p2

  12. A simplified example involving height, d=0: AA AB BB Genotype Genotypic value 160 180 200 Genotype frequency q2 2pq p2

  13. A simplified example involving height, d=0: Situation 1) q=.2 AA AB BB Genotype Genotypic value 160 180 200 Genotype frequency q2 2pq p2

  14. A simplified example involving height, d=0: Situation 1) q=.2 AA AB BB Genotype Genotypic value 160 180 200 Genotype frequency .04 .32 .64

  15. A simplified example involving height, d=0: Situation 1) q=.2 AA AB BB Genotype Genotypic value 160 180 200 Genotype frequency .04 .32 .64

  16. A simplified example involving height, d=0: Situation 1) q=.2 cm AA AB BB Genotype 200 Genotypic value 160 180 200 Genotype frequency .04 .32 .64 180 160

  17. A simplified example involving height, d=0: • Situation 1) q=.2 • Population mean: • = a(p – q) + 2dpq • a = 20 • d=0 • p=.8 • q=.2 • = 12 cm • (on a scale with midpoint=0!) • On actual scale: • = 180cm + 12cm = 192cm a a cm AA AB BB Genotype 200 Genotypic value 160 180 200 Genotype frequency .04 .32 .64 180 160

  18. A simplified example involving height, d=0: • Situation 1) q=.2 • Population mean: • = a(p – q) + 2dpq • a = 20 • d=0 • p=.8 • q=.2 • = 12 cm • (on a scale with midpoint=0!) • On actual scale: • = 180cm + 12cm = 192cm m 192 cm AA AB BB Genotype 200 Genotypic value 160 180 200 Genotype frequency .04 .32 .64 180 160

  19. A simplified example involving height, d=0: • Situation 1) q=.2 • Population mean: • = a(p – q) + 2dpq • a = 20 • d=0 • p=.8 • q=.2 • = 12 cm • (on a scale with midpoint=0!) • On actual scale: • = 180cm + 12cm = 192cm • = a + d (q - p) • = a + 0 • = 20 • aA = -pa = -.8*20 = -16 • aB = qa = .2*20 = 4 m 192 cm AA AB BB Genotype 200 Genotypic value 160 180 200 Genotype frequency .04 .32 .64 180 160

  20. A simplified example involving height, d=0: • Situation 1) q=.2 • Population mean: • = a(p – q) + 2dpq • a = 20 • d=0 • p=.8 • q=.2 • = 12 cm • (on a scale with midpoint=0!) • On actual scale: • = 180cm + 12cm = 192cm • = a + d (q - p) • = a + 0 • = 20 • aA = -pa = -.8*20 = -16 • aB = qa = .2*20 = 4 aA aB m 192 cm AA AB BB Genotype 200 Genotypic value 160 180 200 Genotype frequency .04 .32 .64 180 160

  21. A simplified example involving height, d=0: • Situation 1) q=.2 • a = 20 • d=0 • p=.8 • q=.2 • = 192cm • = 20 • aA = -16 • aB = 4 aA aB m 192 cm AA AB BB Genotype 200 Genotypic value 160 180 200 Genotype frequency .04 .32 .64 180 160

  22. A simplified example involving height, d=0: • Situation 1) q=.2 • a = 20 • d=0 • p=.8 • q=.2 • = 192cm • = 20 • aA = -16 • aB = 4 • bvAA=2aA=-32 • bvAB=aA+aB=-12 • bvBB=2aB=8 bvAA bvBB m 192 cm AA AB BB Genotype 200 Genotypic value 160 180 200 Genotype frequency .04 .32 .64 180 160

  23. A simplified example involving height, d=0: • Situation 1) q=.2 • a = 20 • d=0 • p=.8 • q=.2 • = 192cm • = 20 • aA = -16 • aB = 4 • bvAA=2aA=-32 • bvAB=aA+aB=-12 • bvBB=2aB=8 • So what does this mean? • The mean deviation from the population mean of offspring of short (AA) individuals mated at random with a sample from the population will be -16. The corresponding deviation for offspring of tall (BB) individuals will be only 4. Why? • In random mating with the population, allele frequencies come into play. E.g. if I randomly mate short (AA) individuals with a sample from the population, I will be mating them with only 4% other short individuals, 32% intermediate individuals, and 64% tall individuals. aA aB m 192 cm AA AB BB Genotype 200 Genotypic value 160 180 200 Genotype frequency .04 .32 .64 180 160

  24. +

  25. + = (on average) 176 cm

  26. + Why? Because the short individual is such an outlier (i.e., is so far removed from the mean), and he/she always provides half the genes for the offspring. Thus, the short individual will have a large effect on ‘pulling’ the offspring mean downwards. = (on average) 176 cm

  27. A simplified example involving height, d=0: • Situation 1) q=.2 • a = 20 • d=0 • p=.8 • q=.2 • = 192cm • = 20 • aA = -16 • aB = 4 • bvAA=2aA=-32 • bvAB=aA+aB=-12 • bvBB=2aB=8 • So what does this mean? • The mean deviation from the population mean of AA individuals mated at random with a sample from the population will be -16. The corresponding deviation for BB individuals will be only 4. Why? • In random mating with the population, allele frequencies come into play. E.g. if I randomly mate short (AA) individuals with a sample from the population, I will be mating them with only 4% other short individuals, 32% intermediate individuals, and 64% tall individuals. Still, on average, their offspring will be relatively short (176cm). Why? • Because the alleles of the short individual, being so far removed from the mean, affect the mean genotypic value of the progeny by a lot (by ‘pulling’ it away from the population mean). Thus, allele A is said to have a large (average) effect (aA). aA aB m 192 cm AA AB BB Genotype 200 Genotypic value 160 180 200 Genotype frequency .04 .32 .64 180 160

  28. Back to our example (7.5)…

  29. Situation 1) • aA = -pa = -26.88 -> 174.48-26.88 = 147.68 • aB= qa = 6.72 -> 174.48+6.72 = 181.2 • f(B)=.8 • f(A)=.2 • Situation 2) • aA = -pa = -16.5 -> 154.5-16.5 = 138 • aB= qa = 16.5 -> 154.5+16.5 = 171 • f(B)=.5 • f(A)=.5 • Situation 3) • aA = -pa = -6.48 -> 134.88-6.48 = 128.4 • aB= qa = 25.92 -> 134.88+25.92 = 160.8 • f(B)=.2 • f(A)=.8 AA AA AA AB AB AB BB BB BB Genotype Genotype Genotype Genotypic value Genotypic value Genotypic value 122 122 122 154 154 188 188 188 Genotype frequency Genotype frequency Genotype frequency q2 q2 q2 2pq 2pq 2pq p2 p2 p2 aA aA aB aA aB aB m m m

  30. Situation 1) • aA = -pa = -26.88 -> 174.48-26.88 = 147.68 • aB= qa = 6.72 -> 174.48+6.72 = 181.2 • f(B)=.8 • f(A)=.2 • Situation 2) • aA = -pa = -16.5 -> 154.5-16.5 = 138 • aB= qa = 16.5 -> 154.5+16.5 = 171 • f(B)=.5 • f(A)=.5 • Situation 3) • aA = -pa = -6.48 -> 134.88-6.48 = 128.4 • aB= qa = 25.92 -> 134.88+25.92 = 160.8 • f(B)=.2 • f(A)=.8 • Note that here dominance IS present, so there are deviations from linearity (unlike in the height [d=0] example). However, d is small in the present example (d=-1), so the two situations are fairly similar. In general, however, dominance does complicate things. AA AA AA AB AB AB BB BB BB Genotype Genotype Genotype Genotypic value Genotypic value Genotypic value 122 122 122 154 154 188 188 188 Genotype frequency Genotype frequency Genotype frequency q2 q2 q2 2pq 2pq 2pq p2 p2 p2 aA aA aB aA aB aB m m m

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