QUANTITATIVE GENETICS BPSC 148
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Date Created: 10/29/15
Chapter 13 Maximum Likelihood Estimation of Variances Consider the following three pedigrees families yi yz y3 y4 y5 y6 W The genetic relationships between relatives are different from family to family For example y1 and y2 are fullsibs y3 and y4 are halfsibs ys and y6 are fullsibs but both are cousins of y The number of sibs per family is also different from family to family The analysis of variances method cannot solve such a complicated problem The problem however can be easily solved using the maximum likelihood method Linear models y1 u a1 e1 y2 u a2 e2 y3 u a3 e3 y4 u a4 e4 y5 u a5 as y6 u a6 as y7 u a7 e7 where u population mean xed effect a additive genetic effect of individual 139 with a N N0 VA distribution note VA 6 q environmental error with a e N N 0 VE distribution note V 6 The expectation variance and covariance of the model are Eyl ufori l7 Varyl r1166 rHVA V2 for z39 17 Covylyj G6106Z rUTL1 0VEfor 139j l7 De ne yy1 y7T aa1 a7T ee1 e7T and 11 IT The models can be written in a compact matrix form y1uae y1 1 a1 6 y7 1 a7 6 Now we have Ey In is the expectation vector and Vary V a 7gtlt7 variance covariance matrix of vector y This variance matrix is partitioned into Vary Vara Vare V A634 162 e where OOOOONIHD I OOOOOt NIH IHHOO CO 0000 0000 000 aoooo OOOt IAIquot gl NI D g glw is the additive relationship matrix Note that the diagonal elements of A are all one s because no inbreeding is assumed The offdiagonal elements are zero s for unrelated indiViduals members from different families and nonzero s for genetically related indiViduals Recall that the general form of A is r11 r12 r17 1Fi Z z 2 A r12 r22 r27 Z z 1Fi 2f27 r17 r27 r77 2i7 2f27 1 The actual form of Vary is 10000011000000 100000 0100000 001000 001000 V001000600010006 00001 0000100 00001 0000010 0000 1J 0000001 Chapter 10 Estimating Heritability Part II Full Sib analysis Assume n fullSib families each with k sibs The data structure will be Family Sib 1 Sib 2 Sib k Family mean 1 y11 y12 39 39 39 ylk J71 2 y21 y22 39 39 39 yzk J72 quot ynl ynZ 39 39 39 ynk 7 Covariance between full sibs is C0vFS COVOwyyv WA VD VEC where VEC is called the common environmental variance The total phenotypic variance is VarP VaryU Therefore IMWLhz VE eehz VarP VP 2 VP VP 2 C ovFS if VD and VEC are absent or small The ratio can be estimated as the intraclass P correlation coef cient denoted by t in the analysis of variance ANOVA 121 The intraclass correlation t is similar to the repeatability r Both are intraclass correlation but tis the correlation between different sibs within the same family whereas r is the correlation between different measurements within the same individual Statistical model yy H f WU where yU is the phenotypic value of sibj within the ith family u is the overall mean f is the effect of ith family average of all sibs when the family size is co WU is residual or deviation of yU from the family mean Both and WU are random variables with normal distributions N06 and WU N N0 6 Let us look at the variance ofthe model 2 Z Z VaryUVar Varw1 6 6W 6y V w 4 between family within family The variance of y however is phenotypic variance ie VaryU 6 V The between family variance is equivalent to the covariance between fullsibs ie CovF S 63 This can be shown as follows Let j and j be two sibs within the ith family their phenotypic values are described by yy u f Wt yyv HfzWZJW The variances are 2 2 2 6M Varyy Var Varwy 6f6w and 62 Varyy Var Varwy of 63 M The covariance between yU and yy is C0vylylj C0vu fl WU u fl wt C0v C0v wlj C0vwU ff C0vwlj WV 2 COMM Var 6f Therefore C0vFS CovyUyU 63 We have now associated the covariance between fullsibs with the between family variance Let us compare the statistical model with the genetic model so that an equivalence can be deduced Statistical model 6 6 6 Genetic model VP VA VD VEC VER V E As shown earlier we have the following equivalence 2 6yl3 3 CovFSVAVDVEC c cmVDVEW The intraclass correlation correlation between fullsibs now can be expressed as ratio of variance components 2 Z tzr Covyyylj 6f 2 6f yyl 2 2 2 2 Z Z I GyUGyU 1lof6w 1 6f6w 6f6W Because variance components can be estimated from analysis of variance we have A Z A 6f t A2 A2 6 f 6W When VD and VEC are absent or small we have 112 m 2 The variance components are obtained by the following ANOVA table Source df SS MS EMS Between d Cnl SSf MS 6 k6 Within dfwnk l SSW M SW 5 8 MSW A 1 c 2Msf MSW If the number of siblings per family is different across different families we have unbalanced data Formulas for the SS terms need to be revised and k will be replaced by k0 see Lynch and Walsh 1998 n k2 160 2161 21 1 quot 1 11 2211Cl where IQ is the number of siblings in the 1th family for 139 l n 12 4 Half sib analysis Covariance between halfsibs is C0vHS VA Therefore the intraclass correlation is C0vHS V Lh2 VarP VP 4 The intraclass correlation is similarly estimated via ANOVA by quot2 6 A f t A Z A Z 6 f SW The heritability is now estimated by 22 4 Mixed sibships A common form in which data are obtained with animals is the following A number of males sires are each mated to several females dams the males and females being randomly chosen and randomly mated A number of offspring from each female are measured to provide the data The individuals measured thus form a population of halfsib and fullsib families Assume the population is made of s sires each mated to d dams and each dam reproduces k progenies The total number of individuals is sdk but there are s halfsib families and d fullsib families within each halfsib family The data structure will look like 125 Mixed sibships Sire Dam Sib l Sib k 1 1 y111 yllk 2 y121 ylzk d yldl yldk 2 1 y211 y21k 2 y221 yzzk d yzdl yzak S 1 ysll yslk 2 ySZl 39 39 39 ySZk d ysdl ysdk Nested model of analysis of variances yx H 31 dim WW where Sl N0G is the sire effect dz N N0G is the dam effect and WW N N 063 is the within fullsib family effect The variance components are estimated from the following ANOVA table 126 ANOVA Table for mixed sibship Source df SS MS EMS Between Sire 31 SS MSS a kc dkc Between Dam sd 1 SSd MSd 6 k6 within Sire Between Sib sd c J SSW MSW 6 within Dam You need to equate EMS to the measured MS to solve for the three variance components Figure out by yourselves for each of the three estimated variance components The relationships between the statistical variance components and the genetic components are given below 2 a VlVDlEC 6w VlVDVEW Estimated heritability l Paternal halfsib hA2 46 6 6 6 A 4 A2 2 Maternal halfs1b h2 65 601 6W 26 6 A2 A2 A2 65 6d6w 3 Fullsib i Chapter 17 Multiple Traits and Genetic Correlation Consider two metric charactersX and Y expressed in one individual The phenotypic value of each character can be described by its own linear model PX A X E X PY AY E The phenotypic covariance can be partitioned into genetic and environmental covariances as shown below CovPXPY C0vAX EXAY EY C0vAXAYC0vEXEY CF CA G where C P is called the phenotypic covariance C A is the additive genetic covariance and C E is the environmental covariance In contrast to the environmental covariance between two relatives of the same trait C E always exist because E X and E always happen in the same environment the same individual The partitioning of phenotypic covariance is similar to the partitioning of phenotypic variance De nition of genetic correlation Phenotypic correlation betweenX and Y is de ned as C0VPX9PY CF rP rPXPY GPXGPY GPXGPY The genetic correlation betweenX and Y is de ned by Cow1m CA rA rAXAY 6 XGAY GAXGAY In words genetic correlation is the correlation between the additive genetic effects breeding values of two characters expressed in the same individua The environmental correlation is C0VEX EY CE rErgxgy GEXGEY GEXGEY Let 2 2 e2 SEX l GAX l hz X 2 2 X GPX GPX 2 2 6 6 2 Ey Ay 2 eY 2 l 2 l hY SPY SPY The partitioning of the phenotypic correlation is no FAth rEeXeY which can also be seen from the path diagram Fig 191 Insert Fig 191 here Causes of genetic correlation 1 Pleiotropic effect One gene has effects on both traits This cause is permanent 2 Linkage Genes that control different characters are linked on the same chromosome This is a temporary cause of genetic correlation If linkage is perfect then this cause is of no difference from pleiotropic effect In a population with linkage equilibrium genetic correlation is only caused by pleiotropic effect Cross covariance between relatives We now introduce the concept of cross covariance Cross covariance between two relatives is defined as the covariance between traitX in the first individual and trait Y in the second individual PX2 AX2 EX2 PY2 A2 EY2 PXl AX1 EX1 Individual 1 PYl Al EYl Individual 2 Assume that the two relatives do not share common environmental effect The phenotypic covariance between two relatives is purely caused by their genetic covariance We have COVUJXI PYZ C0VAX1AY2 rC0vAXlAYl rCA Similarly C0VPYl PX2 C0vAYl AX2 rCovAYl AX1 rCA Therefore under the additive model the cross covariance between relatives equals a proportion of the additive genetic covariance between the two traits This proportion is the additive relationship r Estimation of genetic correlation l Sib analysis ANOVA Assume n fullsib families each with k sibs The data structure will be Family Sib 1 Sib 2 Sib k 1 x119y11 xlzaylz xlkaylk 2 leayzi xzzayzz xzkayzk 7 xnl ynl xnz yr xnk ynk In addition to regular analysis of variance for individual traits here we have to include analysis of covariance ANCOVA Following is an ANCOVA table Source of Degree of Sum of cross Mean cross Expected covariation freedom df products products M CP M CP EM CP Between B d n 1 SCPB MCPB SCPBd CW kCB de quotk 7 111de CW 19 4 Let the expected MCP s equal the observed MCP s we have the following system equations CW kCB MCPB CW 008 MCPW The solutions are 68 MCPB MCPW The between family cross covariance component is a proportion of the genetic covariance ie C B rCA In the meantime we have already obtained the variance components of individual traits The genetic correlation is then estimated by f CAXY C300 AXY 8 8 A2 A2 39 Ax Ay GBmGBm The computing formulas for the sum of cross products SCP are n k n k SCPTXY 2206 97 yy 7 22961ij 4107 7 11 11 11 11 SCPBXYki q n ykzm way SCPWXYSCPTXY SCPBXY 1 n k 1k 1 1 k 1k h d E 39 l w ere x quotk 1 Elxy x1 kaU an y quotkg1 yy y k yy 195 2 Parentoffspring correlation path analysis Genetic correlation can be estimated by using path analysis Let X and Y be the phenotypic values of a parent andX and Y are the phenotypic values of a child These phenotypic values are connected through the path diagram shown in Fig 192 The genetic correlation rA is not observable but the cross correlation coefficients can be calculated through data The following estimable phenotypic correlation coefficients are function of heritabilities and genetic correlation 1 2 rXX 7hX i 2 rYY392hY The last two correlation coefficients are called cross correlation coefficients Because they have the same expectation we can combine them together ie i0 rY39X thAlZY We now have three unknowns and three equations rXY39 i172 FY 2 icky rY39X thAhY The estimated genetic correlation is obtained by a rX39YrY39XI Insert Fig 192 here Indirect selection When characters are genetically correlated improvement of the breeding value of characterX can be achieved by selecting the phenotypic value of Y This is called indirect selection In contrast the traditional individual selection ie to select the phenotypic value of traitX to improve the breeding value of traitX is called direct selection The selection response of direct selection is RX inAXPXO AX iXlZXUAX and that of indirect selection is RY inAXPyUAX inAhYO AX The relative ef ciency of indirect selection is de ned as RY erAhYGlx erAhY RX thXGAX thX When the selection intensities are the same for both methods indirect selection is more ef cient that direct selection only if rAhY gt hX In indirectly selection traitX is called the desired character and trait Y is called the secondary character Indirect selection can be considered in the following two situations 1 The desired character X is dif cult to measure 2 The desired character is measured in one seX FA FE AX AY EX EY h hY ex ey PX PY Fig 191 Path diagram of correlation between X and Y 1 1 2 2 m A39x A39y Ax Ay mi hi mi hi X39 Y39 X Y Fig 192 Path diagram of cross correlations DATA Structure for Sib Analysis Family Sib 1 Sib 2 Sib k 1 xnayu xlzaylz xlkaylk 2 x219y21 xzzayzz xzkayzk 7 xnl ynl an yVLZ xnk ynk ANOVA Table for Trait X Source of Degree of Sum of squares Mean squares Expected MS variation freedom df Between B d n l SSBX MSBX SSBXd g oVZWX kogm Within W de nk 7 l SSWX MSWX SSWX d v oVZWX ANOVA Table for Trait Y Source of Degree of Sum of squares Mean squares Expected MS variation freedom df Between B d n l SSBY MSBY SS3Yd 3 oVZW kogm Within W de nk 7 l SSWY MSWY SSWMd v oVZW ANCOVA Table for Traits X and Y Source of Degree of Sum of cross products Mean cross products Expected M CP coVariation freedltln df M CP EM CP Between B d n l SCPB MCPB SCPBd CW kCB Within W de nk 7 l SCPW MCPW SCPWde CW 1910 Chapter 9 Estimating Heritability Part I Heritability narrow sense is de ned as the ratio of the additive genetic variance to the phenotypic variance VA V P W Heritability can also be interpreted as the regression coefficient of breeding value on the phenotypic value CovAP CovAA R CovAA CovA R VarA hi VarP VP VP VP VP 39 AP Therefore heritability can be used to predict breeding value from phenotype A Z bAPP 13 h2P 13 which leads to 21 A7h2P F Heritability is a property not only of a character but also of a population of the environmental circumstance to which the individuals are subjected and of the way in which the phenotype is measured Heritability is de ned as ratio of variance components Therefore we can estimate the variance components first and then convert the variance components into heritability Sometimes heritability can be estimated directly e g the parentoffspring regression analysis There are two types of data used for this purpose data from line crosses and data sampled from outbred population The methods are quite different for different types of data Inbred lines and line crosses Although line crossing experiments are no longer useful for estimating heritability for historical reason it is worthwhile to mention here P1 AA gtlt aa P2 l F1 Aa l 9 F 2 i aa Aa AA 4 Each of the two parental and F1 populations is a genetically homogeneous population Their phenotypic variances within population variances are solely caused by environmental variance ie VarUDl VarPz VarF1 V The phenotypic variance in the F 2 population is caused by both the genetic and environmental variances Therefore we have V0VFz Vs V5 The broad sense heritability can be estimated by 92 VarEVarPl VarPz VarF1l VarF In summary there are four steps in estimating broad sense heritability using inbred lines and line crosses 1 Figure out a population from which the total phenotypic variance can be estimated 2 Figure out a population from which the environmental variance can be estimated 3 Take the difference of the two variances to get an estimated genetic variance and 4 Take the ratio of the genetic variance to the total phenotypic variance to obtain an estimated broad sense heritability We now provide a real life example for estimating the broad sense heritability of early growth of cowpea This was a high school science project conducted by Nicole Xu under the guidance of Dr Jeff Ehlers at UCR in 2006 Table 1 Cowpea height at day 5 measured in centimeter cm for the control and experimental groups RlR15 represent 15 pots in which all plants have the same genotype and G1 G15 represent 15 pots where plants of different pots have different genotypes CONTROL plant1 plant2 plant3 plant4 average R1 90 80 80 95 8625 R2 125 125 110 125 12125 R3 105 90 85 9333 R4 95 105 95 70 9125 R5 80 80 75 80 7875 R6 85 100 85 85 8875 R7 50 110 90 85 8375 R8 95 90 85 85 8875 R9 85 90 95 90 9000 R10 80 70 65 75 7250 R11 80 75 95 100 8750 R12 85 75 90 90 8500 R13 100 95 80 70 8625 R14 70 95 90 8500 R15 75 80 80 80 7875 EXPERIMENTAL plant1 plant2 plant3 plant4 average G1 40 45 115 6667 G2 90 95 85 75 8625 G3 70 95 105 90 9000 G4 65 85 75 70 7375 G5 85 85 85 95 8750 G6 100 75 85 110 9250 G7 100 115 105 115 10875 G8 95 90 100 90 9375 G9 100 110 90 100 10000 G10 95 100 90 105 9750 G11 50 75 75 70 6750 G12 100 95 100 95 9750 G13 120 110 120 120 11750 G14 120 110 105 120 11375 G15 85 90 85 8667 Heights of cowpea plants for the control group all plants have the same genotype and experimental group different genotypes for different pots The means and variances of the 15 pots for the control and experimental groups are given at the bottom of this table Pot CONTROL EXPERIMENTAL 1 8625 6667 2 12125 8625 3 9333 9000 4 9125 7375 5 7875 8750 6 8875 9250 7 8375 10875 8 8875 9375 9 9000 10000 10 7250 9750 11 8750 6750 12 8500 9750 13 8625 11750 14 8500 11375 15 7875 8667 Mean M 87805 91972 Variance V 10949 22803 Data Analysis Let X1X2 X15 be the plant heights ofthe 15 pots for the CONTROL and Y1 Y2 Y15 be the plant heights of the 15 pots for the EXPERIMENTAL The mean and variance of heights across the pots are calculated using the following formulas 115 1 15 M X V X M 2 X 15 i X 1511 X 1 15 1 15 M Y V Y M2 Y 1 5 g 1 Y 15 1 g 1 Y The variance of the control plants VX represents the environmental variance because all plants have the same genotype genetically identical The variance of the experimental plants VY however represents the sum of the genetic variance and the environmental variance Therefore the difference VY V represents the genetic variance The ratio of the genetic variance to the total variance is called the heritability denoted by VY VX V Y H2 Result The mean and the variance of the control plants are MX 87805 and VX 10949 respectively The mean and the variance of the experimental plants are My 91972 and VY 22803 respectively Therefore the environmental variance is 11449 and the genetic variance is 22803 10949 11854 The broad sense heritability is H2 VY VX 22803 10949 11854 VY 22803 22803 about half of the variance of the cowpea growth is controlled by genetic factors and half is controlled by environmental factors Therefore cowpea growth is a highly heritable trait 05198 052 Figure 1 Cowpea heights at day 5 for 15 control plants top panel and 15 experimental plants bottom panel All control plants have the same genotype and each experimental plant has a different genotype CONTROL 14 127 g 10 7 e 8 TS 6 g 47 2 0 1 1 1 1 1 1 1 1 1 1 1 1 1 12 3 4 5 6 7 8 9101112131415 Plant EXPERIMENTAL 14 12 g 10 g 87 i 7 TS 6 i g 47 7 2 i 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 12 3 4 5 6 7 8 9101112131415 Plant Parent Offspring Regression There are two commonly used designs of experiments for estimation of heritability in outbred populations parentoffspring regression and sib analysis We will first discuss the regression approach sib analysis will be described in the next chapter In nonexperimental populations eg human populations all types of relatives are included in data analysis The data structure is complicated so a complicated statistical analysis e g maximum likelihood is required which will be dealt with later Parentoffspring regression analysis is the simplest method There are two parents and one or more sibs in each family and thus there are many different ways to perform the regression analysis Single parent vs single a spring The pair of observation is the phenotypic values of a single parent and a single offspring in each family The data structure looks like Observation Parent Offspring 1 P1 O1 2 P2 02 n Pquot On The phenotypic values of each pair of observations are described by PP AP DP E P0 A0 D0 E0 Assuming that there is no environmental covariance between parent and offspring the phenotypic covariance is equivalent to the genetic covariance The genetic covariance between a parent and an offspring is half the additive genetic variance ie C0VP0PP Vl39 It is known that VarPP 14 ie the variance of the phenotypic value of the parent is the phenotypic variance The regression coefficient of the phenotypic value of an offspring on the phenotypic value of a single parent is C0vPoPp WA lh2 VarPP VP 2 39 b OP This simple relationship allows us to estimate the heritability because the regression coefficient can be estimated from sampled data Therefore M I39d parent vs single offspring The midparent value is de ned as the average of the phenotypic values of the two parents ie 1 P EPfPm The pair of observations are the midparent value and a single offspring in each family The data structure looks like Observation Father Mother Midparent Offspring l P Pml 171 01 2 P Pm 172 02 n Pf Pmquot 17 On The phenotypic covariance between the midparent and an offspring is CovO13 CovOPm Pf CovOPmCovOP VA This is because CovOPm CovOPf K1 The variance of the midparent value is Var13 VarPm Pf 7VarPm VarPf 7VP VP VP Therefore the regression coefficient of the phenotypic value of an offspring on the mid parent value is b Cov0F LVA 2 h2 0P VarP WP This simple relationship allows us to estimate the heritability because the regression coefficient can be estimated from sampled data Therefore Single parent vs mean a spring When there are kprogenies per family we usually take the average of the phenotypic values of the k progenies and regress the mean phenotypic value of the offspring on the phenotypic value of a parent The data structure will look like Observation Parent Mean offspring 1 P1 51 2 P2 52 n Pquot Q k where a 201 is the average of the phenotypic values of k offspring 1 First let us look at the covariance between the mean offspring value and the phenotypic value of the parent k k Cov5P Cov ZOP CovZOP CovOP VA K1 11 1 Therefore Cov5PVA i 51 VarP VP 2 The estimated regression coefficient is A 11 bEP Therefore 119 Chapter 11 Identityby descent Coancestry and Inbreeding Coef cient The coefficient of relationship between two relatives previously denoted by r describes the closedness between two individuals To fully understand the actual meaning of r we must first understand some fundamentally important concepts in population and quantitative genetics that describe the relationship between two genes alleles Allelic variance First let us define the allelic variance and its relationship to the additive genetic variance As described earlier additive genetic variance is the variance of breeding values However breeding value is made of two allelic effects one from the male parent and the other from the female parent Each allele is a haploid ie a single copy of gene The variance of the allelic effect is called the allelic variance To demonstrate the relationship between the allelic variance and the additive genetic variance we must use the following model yuoc5ocd5s where of and ocd are the effects of paternal allele allele from the sire ie father and the maternal allele allele from the dam ie mother respectively 5 is the interaction dominance effect between the two alleles and s is the environmental error Each allelic effect is assumed to be sampled from a normal distribution with mean zero and variance 6 ie 0L50Ld N N06 We may use a genetic notation V 6 for the allelic variance The dominance effect is assumed to be sampled from 5 N N06 distribution Similarly we may use the genetic notations VD 6 and V 6 Under the assumption that the parents are genetically unrelated we have Vary Var0cs Var0cd Var5 Var 2 6y 26 6 6 HF wl VP W Va VE We also know that K V VD It is now obvious that VA 2V ie the additive genetic variance equals twice the allelic variance Genetic covariance between relatives De ne the genetic models of two relatives by G1 A1 D1 ocfocf 81 G2 A2 D2 20L 062i 52 The genetic covariance between 1 and 2 is CovG1 G2 Cov0cf ocf 513 at 52 Cov0cf0cCov0cf0cCov0cf0cCov0cf0cCov5152 Pr0ci a a Pm a a Prof a a Mon 2 a M Pr0ci E 0c Pr0cf E at Pr0cf E at Pr0cf E xi Pm a aProcf a aProcf E aProc E a m quVA UV3 where 0c E at indicate the event that the paternal allele of 1 is identicalbydescent IBD to the maternal allele of 2 Prxf E 0c is the probability that xi E x fu Pr0ci a a Prmi a on Pro a a Prmf a a 1 is called the coancestry between individuals 1 and 2 and um Prmi 2 mm 2 a Prm E a Prmf 2 a2 Pr0ci a aProcf a a mm 2 mm a on is called the fraternity The coefficient of relationship is de ned as twice the coancestry ie r12 2f12 Since 0 S f12 S 1 we have 0 S r12 S 2 which is different from what was previously defined Therefore the covariance between the additive genetic effects of two relatives can be as much as twice the additive genetic variance The fraternity still takes value within the normal range ie 0 S u12 S 1 Let us illustrate the coancestry using the example of fullsibs Sire 141A2 gtlt 143A4 Dam l A1A3A1A4 142143AZA4 We have the following probabilities Ala11 Ais Ag A i A 14 Procisoc 45 Procfzoci 4 0 1 amp5amp amp amp amp amp amp5amp 14 Pr0cfEx 23 0Pr0cfEx fill 4 1 4 A4 AZ 1445A4 Therefore we have 2iG00 i and uuGx mX GXXWXW The additive relationship between fullsibs is i L r12 2f12 2X4 2 Genetic covariance of an individual with itself It is known that the covariance of a variable with itself is always equal to the variance of the variable However this is not necessarily so for the covariance of a breeding value of the same individual CovG1G1 Cov0cf 0cf 510ci 0cf 51 z HA HA A1 D A1 D1 WA r s a s a Cov0c1 0c1 xl 0c1Cov5151 EH EH do as A1 A1 D1 D1 Cov0cf0cf Cov0cfafCov0cf0ciCov0cf0cfo 6 Pr0ci E ocf 6 Pr0cf E Xi6 6 6 2 2 Maj a af no 6 2 2 Pm a af Va VD 1 Pr0cf a af 2V VD HEM VD where F Prm a a1 is called the inbreeding coef cient of individual 1 The covariance of the genetic value of the same individual can be expressed in a general notation C0VG1G1 r1116 11113 1FiVi VD39 Therefore r11 1F1 and M11 1 Because r11 Z l we have the de nition ofthe coancestry of an individual with itself Let us now demonstrate the inbreeding coefficient for individuals resulted from selfing A1 A2 l A1A1A1A2A1A1 The inbreeding coefficient is Ale1 5 As A2 1 EPr0clsocf Ai A 3 1 A1 5 A1 Therefore the coancestry of a selfed individual with itself is f1 1F1 and the coefficient of relationship is r11 fol 1Fi 1 Now let us look at the coancestry between two selfed sibs fu immai a a Pro a on Procf a a Prmf a a 1 l l l l l l 4 2 2 2 2 2 Therefore the covariance between the breeding values of two selfed sibs is equal to the additive genetic variance due to r12 2f12 2 X l Formal de nitions Inbreeding The mating together of individuals that are related to each other by ancestry It is a phenomenon of mating between relatives Identigz by descent Two genes that have originated from the replicate of one single gene in a previous generation Identity by state Two genes that have the same form but originated from different founder alleles Base population Identitybydescent is essentially a comparison between the population in question and some speci c or implied base population In the base population all genes are treated as independent So the previous generation de ned earlier must be a generation at or after the base population Coancestry Coancestry is also called coef cient of kinship It is the probability that two genes taken at random one from each individual are identicalbydescent Coancestry describes the relationships of genes between individuals Inbreeding coe icient The probability that the two genes at a locus in an individual are identicalbydescent Inbreeding coef cient describes the relationship of genes within an individual Per definition ofthe coancestrv and 39 39 139 coef cient the 39 39 139 coefficient of an individual eguals the coancest of the two parents Figurenl 39 betwe identityb 39 39 b state 13s The rst al1e1es ofthe two siblings are IBD because they are copies ofthe second allele omefamen l ct l t are IBS because they are copies of different founder alleles although both have the same onn A1 hld Figure 132 A hypothetical complex pedigree with 22 individuals
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