Lecture 4: Basic Designs for Estimation of Genetic
Lecture 4: Basic Designs for Estimation of Genetic Parameters Heritability Narrow vs. board sense Narrow sense: h2 = VA/VP Slope of midparent-offspring regression (sexual reproduction) Board sense: H2 = VG/VP Slope of a parent - cloned offspring regression (asexual reproduction) When one refers to heritability, the default is narrow-sense, h 2 h2 is the measure of (easily) usable genetic variation under sexual reproduction 2 ( A ; P ) A A r(A ;P )= = = h A PPP Why h2 instead of h? Blame Sewall Wright, who used h to denote the correlation between phenotype and breeding value. Hence, h 2 is the total fraction of phenotypic variance due to breeding values Heritabilities are functions of populations Heritability values only make sense in the content of the population for which it was measured. Heritability measures the standing genetic variation of a population, A zero heritability DOES NOT imply that the trait is not genetically determined Heritabilities are functions of the distribution of environmental values (i.e., the universe of E values) Decreasing VP increases h2. Heritability values measured in one environment (or distribution of environments) may not be valid under another Measures of heritability for lab-reared individuals may be very different from heritability in nature
(A ; P A ) 2 = ) + e = h ( P ) + e p p 2 P 2 2 (e1 = h ) A Heritability and the prediction of breeding values If P denotes an individual's phenotype, then best linear predictor of their breeding value A is The residual variance is also a function of h2: The larger the heritability, the tighter the distribution of true breeding values around the value h2(P - P) predicted by an individuals phenotype. Heritability and population divergence Heritability is a completely unreliable predictor of long-term response Measuring heritability values in two populations that show a difference in their means provides no information on whether the underlying difference is genetic Sample heritabilities People hs Height 0.65 Serum IG 0.45
Back-fat 0.70 Weight gain 0.30 Litter size 0.05 Abdominal Bristles 0.50 Body size 0.40 Ovary size 0.3 Egg production 0.20 Pigs Fruit Flies Traits more closely associated with fitness tend to have lower heritabilities Estimation: One-way ANOVA Simple (balanced) full-sib design: N full-sib families, each with n offspring One-way ANOVA model: zij = + fi + wij 2 f = between-family variance = variance among family mean Trait value in sib jDeviation from Common Effect family of for Mean sib ifamily j from
i =the family mean 2w = within-family deviation varianceof mean of i from tsshe common mean 2P = Total phenotypic variance = 2f + 2w C o v (F u l i )2 S ;fFS b s = ( z ) iw j(2 k [= f)= + w ; ( + f w ) ] i j i k 2 2 2 ) ( ; f + ( w ; ) k i j
i j k 2 + = 4 + A D E c 2 f 2 2 2 2 (1 = + = 4 + ) P A D E c 2 + ( 2 = 4 + ) A D E A D E c 2 )3 ) Covariance between members of the same group equals the variance among (between) groups The variance among family effects equals the covariance between full sibs The within-family variance 2w = 2P - 2f,
N X 2 (X n z ) iiN i= = 1 n 2 (1jz ij i) One-way Anova: N families with n sibs, T = Nn Factor Degrees of freedom, df Sums of Squares (SS) Mean sum of squares (MS) E[ MS ] Among-family N-1 SSf = SSf/(N-1) 2 w Within-family T-N SSW = SSw/(T-N) 2 w +n 2 f M S
f w V a r (w )fz= f2 = n w V (A r2 a f2 )+ + r V (D )2 a w 2 = 4 + E c Estimating the variance components: Since 2Var(f) is an upper bound for the additive variance 2 2 ( M S ) x 2 (V 'ar[(fw M S ) x d f + 2 M S
2 ( M S ) w f w )]F r'n = V a S '2 ( ) n T N + 2 2 2 w + 2 N 1 Assigning standard errors ( = square root of Var) Nice fact: Under normality, the (large-sample) variance for a mean-square is given by ( ( ) ) 2 2 = 4 + V a r ( f ) 1 D E 2 c tSE(h2)F'S=
= h + z 2 z p 2 (1 tFS)[1 + (n 1 )tFS]2 = [N n (1 )] Estimating heritability Hence, h2 < 2 tFS An approximate large-sample standard error for h2 is given by - - - M S 4 5 2 0 f w V rV a (S fa )r = = = 5 n )E w = M S = 2 0 ((z
(h2)'2 V f(10 a r ):4 + r)[1+ V a ((5 w )1 = 2 5 p )0 :4 ]2 = [5 0 (1 )]= 0 :3 1 2 Worked example 10 full-sib families, each with 5 offspring are measured Factor Df SS MS EMS Among-familes 9 SSf = 405 45 2 w Within-families 40 SSw = 800 20 2 w
+5 2f VA < 10 h2 < 2 (5/25) = 0.4 Full sib-half sib design: Nested ANOVA 1 1 n n 2 1 n 2 Full-sibs o* o* o* o* o. * o. * o* o* o. * o o* 1 1 2 2 3 3 .. .. Half-sibs * * k o k
1 2 .. 3 k .. . o* o* o* o* o. * o. * o* o* o. * o* o* o* .. 1 1 2 2 3 3 k .. k 1 2 .. 3 k s Estimation: Nested ANOVA Balanced full-sib / half-sib design: N males (sires) are crossed to M dams each of which has n offspring Nested ANOVA model: zijk = + si + dij+ wijk
= between-sire variance = variance in sire family means Effect Effect of sire of dam Within-family i = deviation j of sire i deviation = deviation of kth Value of the kth offspring Overall mean 2 variance within sires = and of mean of dams mean of iisof offspring family dam j from from fromsire the mean overall of the the kthamong dam for sire d =from overall mean mean ij-th same family sire ariance of dam means for the 2w = within-family variance 2 T = 2 s + 2 d + 2 w N M i X 2 2 2 2 +
n M n M n ( z ) i w d s iX jiN = 1 N M 2 2 + n 2 w d i j i i= j1 = 1 M n w 2 (jk= z ij) i1 jk Nested Anova: N sires crossed to M dams, each with n sibs, T = NMn Factor Df SS MS Sires N-1 SSs = SSs/(N-1)
Dams(Sires) N(M-1) SSs = SSd /(N[M-1]) Sibs(Dams) T-NM SSw = SSw/(T-NM) EMS M S s d V a r ( s ) = n d w d e w 2 2 2 = d z s w (F S )(P H S ) Estimation of sire, dam, and family variances: Translating these into the desired variance components Var(Total) = Var(between FS families) + Var(Within FS) 2w = 2z - Cov(FS)
Var(Sires) = Cov(Paternal half-sibs) 2 2 2 2 = (w = P H S ) d z s w s ( F S ) ( P H S ) F z 2 2 A '2w s 4 2 D 2 + d E c 2 3 A s 2 4 Summarizing, Expressing these in terms of the genetic and environmental variances ,
C o v ( P H S ) V a r ( s ) tPH = = S V a r z z F + d F Intraclass correlations and estimating heritability 4tPHS = h2 h2 < 2tFS Note that 4tPHS = 2tFS implies no dominance or shared family environmental effects 2 2 2 + 1 0 3 w s d 2 + w d 2 2 = M S =
2 0 w w 2 2 M S 1 7 0 2 = 4 0 d w A s = = 1 5 d n 4 s d 0 s N 3 2 4 0 2 2 2 A 2 + 5 h = = : 8 9 P w 5 P 2 2 2 =
1 5 (0 = 4 )A2+ (1 )D = 4 + d D E c 2 2 + = 2 0 E E c cD Worked Example: N=10 sires, M = 3 dams, n = 10 sibs/dam Factor Df SS MS Sires 9 4,230 470 Dams(Sires) 20 3,400 170 Within Dams 270 5,400 20
EMS z = + b ( z ) + e o p i o j iE i 2 2 ( z ; ) ( = 2 ) + ( E ; ) h ( E ; ) o p o p o p A ( b )2 '= = = + je o p
2 2 2 2 z z 2 h 1 2 z Parent-offspring regression Single parent - offspring regression The expected slope of this regression is: Residual error variance (spread around expected values) Shared environmental values To avoid this term, typically regressions are male-offspring, as female-offspring more likely to share environmental values z + m i f z = + b + e ie j2 o o M P i 2 C o v [ z ; (
+ z ) = 2 ] o m f b = o k M P V a r 2 C o v ( ; z ) = 2 h o f = 1 2 V a r 4 o p = 2 b (z )jp Midparent - offspring regression The expected slope of this regression is h2, as Residual error variance (spread around expected values) Key: Var(MP) = (1/2)Var(P)
Hence, even when heritability is one, there is considerable spread of the true offspring values about their midparent values, with Var = VA/2, the segregation variance 2 n ( t b ) + ( 1 t ) o j p 'a V a r ( b ) jV o p N n 8 2 2 t = h 4 f o r h a l s i b h = V a r 2 = 4 V a r b
H S < j p 2 2 = + D E : 2 u F z 2 2 [ n ( t b = 2 ) + ( 1 t ) ] F S F S o j M P 2 r(h )V a r(b )'N o jM P n c Standard errors Single parent-offspring regression, N parents, each with n offspring Squared regression slope Sib correlation
Total number of offspring Midparent-offspring regression, N sets of parents, each with n offspring Midparent-offspring variance half that of single parent-offspring variance = h V a r ( z ) n 0 2 2 h = ( b ) m i n o j M P A l 2 V a r ( z ) C o v ( A ) V a r ( z ) n l ; n n 0 2
2 (b )ojM = P n V a r z A lA l A p n Estimating Heritability in Natural Populations A lowersibs bound can be in placed of heritability using Often, are reared a laboratory environment, parents parent-offspring from nature andregressions their lab-reared offspring, making and sib ANOVA problematic for estimating heritability Why is this a lower bound? where Covariance between breeding value in nature and BV in lab is the additive genetic covariance between environments and hence 2 < 1
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