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Course: Advanced Animal Breeding

Course: Advanced Animal Breeding

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Course: Advanced Animal Breeding

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  1. Course: Advanced Animal Breeding MS program in Animal Production Faculty of Graduate Studies An-najah National University Instructor: Dr. Jihad Abdallah Mean and genetic variance of quantitative traits

  2. Simply Inherited and Polygenic Traits • Simply inherited traits: traits affected by one or few genes (coat color, presence of horns, genetic defects like spider syndrome in sheep). • Phenotypes of simply inherited traits are placed into categories (qualitative or categorical traits) • Very little affected by the environment.

  3. Polygenic traits: are traits affected by many genes (no single gene having an overriding effect) like growth rate, milk production, birth weight, etc. • Generally described in numbers. • Typically quantitative or continuous in their expression (an exception is Dystocia which is affected by many genes but pohenotype is described in categories) • Polygenic traits are affected by the environment.

  4. Different breeding approaches are used to improve simply inherited and polygenic traits. This is due to the different number of genes involved in each type. • For polygenic traits, it is more difficult to observe the effects of individual genes and search specific information on each one (we do not try to identify the genotype on many loci), instead we try to identify the net effect of the many genes the individual have(i.e, quantify the genetic merit of the individual) • For simply inherited traits we search to identify specific genotypes using test mating (or test cross) and molecular techniques.

  5. The mean and genetic variance for Quantitative Traits

  6. Model of Falconer and Mackay (1996)for a Single Locus • P = G + E • = A + D + E • P = phenotypic value • G = Genotypic value (influence of the genotype) • A = Breeding value • D = dominance deviation • E = Environmental deviation • The mean of E is taken as zero such that the mean phenotypic value (mean of P) is equal to the mean genotypic value (mean of G)

  7. Consider a single locus with two alleles (A1 and A2) with A1 being the desirable allele (the allele which increases the value). We have three genotypes with their scaled genotypic values and frequencies: Genotypescaled genotypic valuegenotypic frequency A1A1 +a p2 A1A2 d 2pq A2A2 -a q2 The 0 value is the midpoint between the genotypic values of both homozygote genotypes midpoint A2A2 A1A2 A1A1 -a 0 d +a The range is 2a in absence of overdominance

  8. The value d of the heterozygote depends on the degree of dominance: No dominance  d = 0 A1 is dominant over A2  d is positive A2 is dominant over A1  d is negative Complete dominance  d = +a (if A1 is dominant) or d = -a (if A2 is dominant) Overdominance  d > +a ( A1 is dominant over A2)  d < -a ( A2 is dominant over A1) Degree of dominance = d/a

  9. Example: pygmy gene in mice • Genotype + + + pg pg pg • Genotypic value 14 12 6 • Midpoint = (14 + 6)/2 =10 • a = 14 – 10 = 4 g • d = 12 – 10 = 2 g • Degree of dominance = 2/4= 0.5

  10. Population mean (From Falconer and Mackay, 1996)

  11. M = a (p – q) + 2dpq • This is both the mean genotypic value and the mean phenotypic value of the population expressed as a deviation from the midpoint value. Note that the mean of environmental deviations is 0. • For the previous example, if p = 0.9 (q = 0.1) then: • M = 4(0.9 – 0.1) +2 (2 x 0.9 x 0.1) = 3.56 • The actual population mean = 10 + 3.56 = 13.56

  12. The average effect of an allele • Parents pass on genes to their offspring and not genotypes. Therefore, we need to consider the average effect of alleles as well. • The average effect of an allele is defined as: the mean deviation from the population mean of the genotypic values of individuals which received this allele from one parent and received from the other parent has come at random from the population. Average effect of A1 Average effect of A2

  13. (From Falconer and Mackay, 1996)

  14. The average effect of the gene substitution • The average effect of gene substitution is the difference between the average effects of the two alleles. • It can be viewed as the effect of substituting one allele in the population with the other allele. • When A2 genes are chosen at random, a proportion p will be found in A1A2 genotypes (p is the frequency of A1) and a proportion q in A2A2. • Changing A1A2 to A1A1 changes the value from d to +a  the effect of substitution is (a – d) • Changing A2A2 to A1A2 changes the value from -a to d  the effect of substitution is (d + a)

  15. The average change (the average effect of gene substitution) is therefore: The average effects of alleles can be expresses in terms of the average effect of gene substitution:

  16. Breeding value • Because parents pass on genes and not genotypes to their progeny, it is the average effects of the parents’ genes that determine the mean genotypic value of the progeny. • If an individual is mated to a number of individuals taken at random from the population, then its breeding value is twice the mean deviation of the progeny from the population mean. • Defined in terms of average effects, the breeding value of an individual is equal to the sum of the average effects of all the alleles it carries on all loci.

  17. For a single locus with two alleles, the breeding values of the genotypes are: GenotypeBreeding value A1A1 A1A2 A2A2 These breeding values are expressed as deviations from the population mean  the mean of breeding values in the population is equal to 0 NOTE:  = Breeding valueA1A1 – Breeding valueA1A2 = Breeding ValueA1A2 – Breeding ValueA2A2

  18. Pygmy Example For q = 0.1 Average of breeding values = (0.9)2 (0.48) + (2 x 0.1 x 0.9) (-1.92) + (0.1)2 (- 4.32) = 0 Similarly, the average of breeding values for q = 0.4 is 0

  19. The breeding value is also referred to as the additive genetic value because the variation in breeding values is due to the additive effects of the alleles. • Because the breeding value expresses the value transmitted from parents to offspring and each parent passes half of its genes to its offspring, the expected value (average) of the breeding values of the offspring is the average of the breeding values of the two parents: O = offspring, S = sire, D = dam

  20. Dominance Deviation • Dominance deviation is due to interaction between alleles at the same locus (within locus interactions) • G = A + D

  21. Additive Dominance Fitting additive effect only • = 2.40 M + Breeding Value of aa = 13.56 – 4.32 Breeding value = Intercept + b1 (AL) Breeding value AA = 9.24 + 2.4(2) = 14.04  14.04 – 13.56 = 0.48 (or 2x 0.1 x 2.4) Breeding value Aa = 9.24 + 2.4(1) =11.64  11.64 – 13.56 = -1.92 Breeding value aa = 9.24 + 2.4(0)=9.24  9.24 – 13.56 = -4.32

  22. Fitting full model (additive + dominance effects) Dominance = d = a Additive effect Genotypic value = Intercept + b1(AL) + b2(AQ) GAA = 6 + 4(2) + 2 (0) = 14  14– 13.56 = +0.44 GAa = 6 + 4 (1) + 2 (1) = 12  12 – 13.56 = -1.56 Gaa = 6 + 4 (0) + 2 (0) = 6  6 – 13.56 = -7.56

  23. Interaction Deviation (Epistasis)

  24. The variance of a quantitative trait Components of the phenotypic variance:

  25. Additive and Dominance variance

  26. Complete dominance No dominance VG VA VD Overdominance

  27. More than one locus • When two or more loci are considered, epistatic effects may be important and contribute to the genetic variance. • As we have seen earlier, epistasis is the interaction between alleles at different loci. • Interaction can be subdivided according to the number of loci involved: two-factor interaction, three-factor interaction, and so on. • Two-factor interaction: arises from interaction between alleles on two loci • Three-factor interaction: arises from interaction between alleles on three loci. • Four-factor and higher order interactions contribute little to the variance and thus can be ignored.

  28. The next subdivision of the interaction variance is based on whether the interaction involves breeding values or dominance deviations. Two-factor interactions have three sorts: • Additive x additive variance (VAA): due to interaction between the two breeding values (linear x linear) • Additive x Dominance variance (VAD): due to interaction between the breeding value of one locus and the dominance deviation of the other locus (linear x quadratic). • Dominance x Dominance variance (VDD): due to interaction between the dominance deviations of both loci (quadratic x quadratic). The interaction variance is divided into the following components:

  29. The total genetic variance is thus divided into: • Additive variance • Dominance variance • Interaction (Epistatic) variance • VG = VA + VD + VI VP = VA + VD + VI + VE • This includes some simplifying assumptions: - no interaction between the genotype and the environment - no correlation between genotype and the environmental deviation (good genotypes are not given a better environment) - The population is in equilibrium under random mating

  30. Example (Jalal and Karam, 2003): two loci with pA = 0.4 and pB = 0.8 Decompose the total genetic variance into its components (additive, dominance and epistasis).

  31. Compute sum of squares

  32. Locus A: p = 0.4, q = 0.6 AAAaaa • Genotypic value: 47.6 45.92 24.8 • Frequency: 0.16 0.48 0.36 • Midpoint = (47.6+24.8)/2 =36.2 • a = 47.6 –36.2=11.4 • d = 45.92 – 36.2 = 9.72 • VA = 2pq[a + d(q-p)]2 = 2(0.4)(0.6)[11.4+9.72(0.6-0.4)]2 = 0.48[13.344]2 = 85.46 • VD = [2pqd]2 = [2(0.4)(0.6)(9.72)]2 = 21.77 • Total for locus A= 85.46+21.77=107.23

  33. Locus B: p = 0.8, q = 0.2 BBBabb • Genotypic value: 41.0 35.4 25.44 • Frequency: 0.64 0.32 0.04 • Midpoint = (41.0+25.44)/2 =33.22 • a = 41 –33.22=7.78 • d = 35.4 – 33.22 = 2.18 • VA = 2pq[a + d(q-p)]2 = 2(0.8)(0.2)[7.78+2.18(0.2-0.8)]2 = 0.32[6.472]2 = 13.40 • VD = [2pqd]2 = [2(0.8)(0.2)(2.18)]2 = 0.49 • Total for locus B = 13.4038+0.4866=13.89

  34. Partitioning the total variation

  35. Solution using regression analysis (SAS) data var_comp; input genotype$ freq gen_value AL AQ BL BQ ALBL ALBQ AQBL AQBQ@; cards; AABB 0.1024 50 2 0 2 0 4 0 0 0 AaBB 0.3072 50 1 1 2 0 2 0 2 0 AABb 0.0512 45 2 0 1 1 2 2 0 0 AaBb 0.1536 40 1 1 1 1 1 1 1 1 AAbb 0.0064 30 2 0 0 0 0 0 0 0 aaBB 0.2304 25 0 0 2 0 0 0 0 0 Aabb 0.0192 28 1 1 0 0 0 0 0 0 aaBb 0.1152 25 0 0 1 1 0 0 0 0 aabb 0.0144 20 0 0 0 0 0 0 0 0 ; procregdata=var_comp; model gen_value=AL BL/ss1; weight freq; run; procregdata=var_comp; model gen_value=AL BL AQ BQ/ss1; weight freq; run; procregdata=var_comp; model gen_value=AL BL AQ BQ ALBL ALBQ AQBL AQBQ/ss1; weight freq; run;

  36. Number of copies of B allele Number of copies of A allele (linear “additive” effect of locus A) 1 if Aa and 0 otherwise (quadratic “Dominance” effect of locus A) 1 if Bb and 0 otherwise (quadratic “dominance” Effect of locus B)

  37. Fitting additive effects only: Breeding value of aabb  for locus A  for locus B

  38. Fitting additive and dominance effects:

  39. Fitting full model (additive + dominance + interactions):

  40. Additive variance Dominance variance (A x A) (A x D) Interaction (epistatic variance) (D x A) (D x D)

  41. data var_comp; input genotype$ freq gen_value AL AQ BL BQ ALBL ALBQ AQBL AQBQ@; cards; AABB 0.1024 50 2 0 2 0 4 0 0 0 AaBB 0.3072 50 1 1 2 0 2 0 2 0 AABb 0.0512 45 2 0 1 1 2 2 0 0 AaBb 0.1536 40 1 1 1 1 1 1 1 1 AAbb 0.0064 30 2 0 0 0 0 0 0 0 aaBB 0.2304 25 0 0 2 0 0 0 0 0 Aabb 0.0192 28 1 1 0 0 0 0 0 0 aaBb 0.1152 25 0 0 1 1 0 0 0 0 aabb 0.0144 20 0 0 0 0 0 0 0 0 ; procregdata=var_comp; model gen_value=AL BL/ss1; weight freq; run; procregdata=var_comp; model gen_value=AL BL AQ BQ/ss1; weight freq; run; procregdata=var_comp; model gen_value=AL BL AQ BQ ALBL ALBQ AQBL AQBQ/ss1; weight freq;run; Quit;