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Reminder - Means, Variances and Covariances. Covariance Algebra.

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Presentation on theme: "Reminder - Means, Variances and Covariances. Covariance Algebra."— Presentation transcript:

1 Reminder - Means, Variances and Covariances

2 Covariance Algebra

3 Covariance and Correlation Correlation is covariance scaled to range [-1,1]. For two traits with the same variance: Cov(X 1,X 2 ) = r 12 Var(X)

4 Phenotypes, Genotypes and environment A phenotype (P) is composed of genotypic values (G) and environmental deviations (E): P = G + E Whether we focus on mean, variance, or covariance, inference always comes from the measurement of the phenotype A distinction will be made: V will be used to indicate inferred components of variance   2 will be used to indicate observational components of variance Mean genotypic value is equal to the mean phenotypic value Genotypic values are expressed as deviations from the mid- homozygote point E(G j ) =  P j )=  Pj

5 Genotypic values Consider two alleles, A 1 and A 2, at a single locus. The two homozygous classes, A 1 A 1 and A 2 A 2, are assigned genotypic values +a and -a, respectively. Assume that the A 1 allele increases the value of a phenotype while the A 2 allele decreases the value. The heterozygous class, A 1 A 2, is assigned a genotypic value of d Zero is midpoint between the two genotypic values of A 1 A 1 and A 2 A 2 ; d is measured as a deviation from this midpoint

6 A1A1A1A1 A2A2A2A2 A1A2A1A2 -a-a+a+a0d Genotype Genotypic value

7 Properties of the Genotypic Values and Environmental Deviations The mean of environmental deviations is zero E j = P j – G j,, (  E = 0) The correlation between genotypic values and environmental deviations for a population of subjects is zero (  GE =.00)

8 Elements of a population mean

9 Population mean  P =  G =  G k p k Multiply frequency by genotypic value and sum Recall that p 2 - q 2 = (p + q)(p - q) = p - q  P = a(p - q) + 2dpq

10 Additive model Assume a A and a B correspond to A 1 A 1 and B 1 B 1 A 1 A 1 B 1 B 1 = a A + a B So that  P =  a(p - q) + 2  dpq

11 Average effect for an allele (  ) Population properties vis a vis family structure Transmission from parent to offspring; parents pass on genes and not genotypes Average effect of a particular gene (allele) is the mean deviation from the population mean of individuals which received that gene from one parent (assuming the gene transmitted from the other parent having come at random from the population)

12 Average effect for an allele (  )

13 Thus, the average effect for each allele also can be calculated for A 1 and A 2 in the following manner (Falconer, 1989):  1 = pa + qd - [a(p - q) + 2dpq] and  2 = -p[a + d(q - p)]

14 Average effect of a gene substitution Assume two alleles at a locus Select A 2 genes at random from population; p in A 1 A 2 and q in A 2 A 2 A 1 A 2 to A 1 A 1 corresponds to a change of d to +a, i.e., (a - d); A 2 A 2 to A 1 A 2 corresponds to a change of -a to d, (d + a) On average, p(a - d) plus q(d + a) or  = a + d(q - p)

15 When gene frequency is greater  is greater

16 Breeding Value (A) The average effects of the parents’ genes determine the mean genotypic value of its progeny Average effect can not be measured (gene substitution), while breeding value can Breeding value: Value of individual compared to mean value of its progeny Mate with a number of random partners; breeding value equals twice the mean deviation of the progeny from the population mean (provides only half the genes) Breeding value is interpretable only when we know in which population the individual is to be mated

17 Breeding Value Genotype: Breeding value A 1 A 1 : 2  1 = 2q  A 1 A 2 :  1 +  2 = (q - p)  A 2 A 2 : 2  2 = -2p  Mean breeding value under HWC equilibrium is zero 2p 2 q  + 2pq(q - p)  - 2q 2 p  which equals... 2pq  (p + q - p - q) = 0

18 Dominance deviation Breeding values are referred to as “additive genotype”; variation due to additive effects of genes A symbolizes the breeding value of an individual Proportion of  2 P attributable to  2 A is called heritability (h 2 ) G = A + D Statistically speaking, within-locus interaction Non-additive, within-locus effect A parent can not individually transmit dominance effects; it requires the gametic contribution of both parents

19 Genotypic values, breeding values, and dominance deviation +a+a d 0 -a-a -2p  (q - p)  0 2q2q A1A1A1A1 A1A2A1A2 A2A2A2A2 }} Genotypic values Breeding values 2pqq2q2 p2p2

20 Genotypic values, breeding values, and dominance deviation Regression of genotypic value on gene dosage yields the genotypic values predicted by gene dosage average effect of an allele that which “breeds true” If there is dominance, this prediction of genotypic values from gene dosage will be slightly off dominance is deviation from the regression line

21 Epitasis - Separate analysis locus A shows an association with the trait locus B appears unrelated Locus A Locus B

22 Epitasis - Joint analysis locus B modifies the effects of locus A

23 Genotypic Means Locus A Locus BAAAaaa BB  AABB  AaBB  aaBB  BB Bb  AABb  AaBb  aaBb  Bb bb  Aabb  Aabb  aabb  bb  AA  Aa  aa 

24 Partitioning of effects Locus A Locus B MP MP

25 4 main effects M P M P Additive effects

26 6 twoway interactions M PM P   Additive-additive epistasis M PP M  

27 4 threeway interactions MPM P  M P MP   M P  MP   Additive- dominance epistasis

28 1 fourway interaction MMP  Dominance- dominance epistasis P

29 Two loci AAAaaa BB Bb bb m m m m m m m m m + a A – a A + d A + a B – a B + d B – aa + aa + dd+ ad – da + da – ad

30 Covariance matrix Sib 1Sib 2 Sib 1  2 A +  2 D +  2 S +  2 N  2 A + z  2 D +  2 S Sib 2  2 A + z  2 D +  2 S  2 A +  2 D +  2 S +  2 N Sib 1Sib 2 Sib 1  2 A +  2 D +  2 S +  2 N ½  2 A + ¼  2 D +  2 S Sib 2 ½  2 A + ¼  2 D +  2 S  2 A +  2 D +  2 S +  2 N

31 Detecting epistasis  The test for epistasis is based on the difference in fit between - a model with single locus effects and epistatic effects and - a model with only single locus effects,  Enables us to investigate the power of the variance components method to detect epistasis

32 AB Y  a b True Model A Y  a* Assumed Model a* is the apparent co-efficient a* will deviate from a to the extent that A and B are correlated

33 Phenotypic variance Again, assume P = G + E Thus differences in phenotypes, measured as variance and symbolized as V P, can be decomposed into both genetic and environmental variation, V G and V E, respectively. V P = V G + V E V G is comprised of three kinds of distinct variance: additive (V A ), dominant (V D ), and epistatic (V I ). V P = (V A + V D + V I ) + V E

34 Analysis of variance

35 Additive (V A ) and dominance variance (V D ) The covariance between breeding values and dominance deviations equals zero so that V G = V A + V D + V I V A = 2pq[a + d(q - p)] 2 V D = d 2 (4q 4 p 2 + 8p 3 q 3 + 4p 4 q 2 ) = (2pqd) 2

36 Additive and dominance variance If d = 0, then V A = 2pqa 2, where q is the recessive allele If d = a, then V A = 8pq 3 a 2 If p = q =.50 (e.g., cross of two inbred strains) V A = 1/2a 2 V D = 1/4d 2 In general, genes at intermediate frequency contribute more variance than high or low frequencies

37 Epistatic variance (V I ) Epistatic variance beyond three or more loci do not contribute substantially to total variance Three types of two-factor interactions (breeding values by dominance deviations) additive x additive (V AA ) additive x dominance (V AD ) dominance x dominance (V DD )

38 Environmental variance Special environmental variance (V Es ) within-individual component temporary or localized circumstance General environmental variance (V Eg ) between-individual component permanent or non-localized circumstances Ratio of between-individual to total phenotypic is an intraclass correlation (r)

39 Summary of variance partitioning

40 Components of variance - Summary Phenotypic Variance EnvironmentalGeneticGxE interaction and correlation

41 Components of variance - Summary Phenotypic Variance EnvironmentalGeneticGxE interaction Additive DominanceEpistasis and correlation

42 Components of variance - Summary Phenotypic Variance EnvironmentalGenetic Additive DominanceEpistasis Quantitative trait loci GxE interaction and correlation

43 Resemblance of relatives Degree of relative resemblance is a function of additive variance, i.e, breeding values The proportionate amount of additive variance is an estimate of heritability (V A / V P ) Intraclass correlation coefficient t =  2 B /  2 B +  2 W Between and within full-sibships, for example

44 Resemblance of relatives b OP = Cov OP /  2 P New property of the population is covariance of related individuals

45 Cross-Products of Deviations for Pairs of Relatives AAAaaa AA(a-m) 2 Aa(a-m)(d-m)(d-m) 2 aa(a-m)(-a-m) (-a-m)(d-m)(-a-m) 2 The covariance between relatives of a certain class is the weighted average of these cross-products, where each cross-product is weighted by its frequency in that class.

46 Offspring and one parent Individual genotypic values and those of their offspring produced by random mating When expressed as normal deviations, the mean value of the offspring is 1/2 the breeding value of the parent Covariance between individual’s genotypic value (G) with 1/2 its breeding value (A)

47 Covariance for Parent-offspring (P-O) AAAaaa AAp 3 Aap 2 qpq aa0 pq 2 q 3 Covariance = (a-m) 2 p 3 + (d-m) 2 pq + (-a-m) 2 q 3 + (a-m)(d-m)2p 2 q + (-a-m)(d-m)2pq 2 = pq[a+(q-p)d] 2 = V A / 2

48 Offspring and one parent G = A + D so that covariance is between (A + D) and 1/2A; sum of cross products equal  1/2A(A + D) = 1/2  A 2 + 1/2  AD Cov OP = (1/2  A 2 + 1/2  AD) / # of parents Recall that Cov AD = 0 Cov OP = 1/2V A (i.e., 1/2 the variance of breeding values)

49 Offspring and one parent: Effects of a single locus

50 Offspring and one parent Mean genotypic values of the offspring are 1/2A of the parents Mean cross product equals Frequency X Genotypic value of the parent X Mean genotypic value of the offspring Cov OP = pq  2 (p 2 +2pq+q 2 )+2p 2 q 2  d(-q+q-p+p)=pq  2 = 1/2V A (Note: V A = 2pq  2 )

51 Covariance of MZ Twins AAAaaa AAp 2 Aa02pq aa00q 2 Covariance = (a-m) 2 p 2 + (d-m) 2 2pq + (-a-m) 2 q 2 = 2pq[a+(q-p)d] 2 + (2pqd) 2 = V A + V D

52 Twins Dizygotic twins are fulls sibs and their genetic covariance is that of full sibs Monozygotic twins have identical genotypes, i.e., no genetic variance within pairs so that Cov (MZ) = V G

53 Covariance for Unrelated Pairs (U) AAAaaa AAp 4 Aa2p 3 q4p 2 q 2 aap 2 q 2 2pq 3 q 4 Covariance = (a-m) 2 p 4 + (d-m) 2 4p 2 q 2 + (-a-m) 2 q 4 + (a-m)(d-m)4p 3 q + (-a-m)(d-m)4pq 3 + (a-m)(-a-m)2p 2 q 2 = 0

54 Resemblance in general Let r be the fraction of V A and u the fraction of V D so that Cov = rV A + uV D P, Q two individual in relationship with parents A,B and C,D and f coancestry r = 2f PQ and u = f AC f BD + f AD f BC For inbred relatives, r = 2f PQ / [(1 + F P )(1 + F Q )] 1/2

55 Resemblance in general Coefficient r of the additive variance is sometimes called the coefficient of relationship (the correlation between the breeding values A) Coefficient u represents the probability of the relatives having the same genotype through identity by descent It is zero unless the related individuals have paths of coancestry through both of their respective parents, (e.g., full sibs and double first cousins)

56 Environmental covariance V E = V Ec + V Ew V Ec ; common, i.e., contributes to variance between means of families but not the variance within (covariance among related individuals) V Ew ; within, i.e., arises from independent of coefficient of relationship Maternal effects and competition

57 Phenotypic resemblance between relatives

58 Heritability Regression of breeding value on phenotypic value Index of response to genetic selection Estimated with offspring-parent regression, sib analysis, intra-sire regression of offspring on dam, or combined estimates plus other methods (Markel et al., 1995, 1999)

59 Heritability: Ratio of additive genetic variance to phenotypic variance h 2 = V A / V P Regression of breeding value on phenotypic value h 2 = b AP r AP = b AP  P /  A = h 2 (1/h) = h

60 Heritability: Twins and human data

61

62 C = B 1 P + B 2  + B 3 P  P  PP B1B1 B2B2 B3B3 C

63 A1A1 C1C1 E1E1 D1D1 A2A2 C2C2 E2E2 D2D2 P1P1 P2P2 Twin 1 Twin 2 acedaced 1.0 (MZ) or.25 (DZ) 1.0 (MZT,DZT) or 0.0 (MZA, DZA) 1.0 (MZ) or.5 (DZ)


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