QUANTITATIVE GENETICS BPSC 148
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Date Created: 10/29/15
Chapter 14 Basic Concepts of Selection Evolutionary forces Evolutionary forces are factors that can change the genetic properties of a population There are four evolutionary forces selection mutation migration and genetic drift Selection is considered to be the most important one Selection is de ned as a phenomenon that parents who contribute to the next generation are not a random sample but selected based on some criteria Genetic properties of a population include gene frequencies and genotype frequencies However these are not observable what we can observe are the population mean and variance that re ect gene and genotype frequencies of the population Change in gene and genotype frequencies Here we use deleterious recessive gene as an example Assuming that Al is completely dominant overAz allele so that A1141 and11A are indistinguishable in terms of selection Genotype 141A1 141A2 A2 A2 Total Initial freq p2 2pq q2 1 Coef cient of 0 0 3 Selection Fitness 1 1 ls Garnetic p2 qu 13q2 13q2 contribution The tness denoted by w is de ned as w 1 s where s is the selection coef cient The garnetic contribution of each genotype is the initial frequency multiplied by the tness After selection the genotype frequencies will change ie the genotype frequencies in the selected parents will be different from the frequencies of the initial unselected population Genotype 141A1 141A2 A2 A2 Total New freq in p2 2 pg 1 3M2 1 selected parents l sq2 1 59 l sq2 De ne the initial frequency of A1 and A2 by p0 and q0 respectively then the new allelic frequencies forll and A2 are 2 2 p1PlH1 p0 p0q0 p0 1 qu 2 1 qu l sqg lW2 1 Mar 617 sq2 q1Q1H1 1 20 0t 20 20 sq0 2 1 qu 1 qu The change of gene frequency in one generation of selection is 2393090 1 M q1 qo M 0 S 0 So ql S qo meaning that frequency of A2 has been decreased due to selection against this allele If the selected parents are randomly mated the genotype frequencies in the offspring will be Genotype Frequency A1141 1 11 2 A1142 2q11q1 Aw q Therefore after one generation of selection both the gene and genotype frequencies have changed As a result the population mean and genetic variance have also changed 162 Changes of a metric character Truncation selection An individual is selected if its phenotypic value is greater than a threshold T otherwise it is culled Insert Fig 171 here The tness of an individual is de ned as 1 if y 2 T 0 if y lt T Selection dz erentials De ne 7 and J75 as the means of the parental population before and after selection respectively The selection differential is de ned by S J73 y Response to selection Response to selection is de ned as the difference between the mean of offspring and the mean of the parental generations Denote the response to selection by S the relationship between R and S is given by R hZS This equation also called the selection equation is important in plant and animal breeding as well as in evolution Selection intensity Selection differentials cannot be compared for different traits A large value of selection differential does not mean a strong selection The strength of a selection is measured by selection intensity Selection intensity denoted by i is also called standardized selection differential It is defined as the selection differential divided by the phenotypic standard deviation ie where UP is the standard deviation of phenotypic value ie 6P 1 6 46 JV We now express the selection response as a function of the selection intensity RhZSh2ioP hioA Proportion selected p The proportion selected p is the ratio of the number of selected individuals to the total number of individuals measured Theoretically if y is normally distributed ie y N NL6p is defined by T TLUP t p 1Lofydy 1 Lo fy H6pdy H6P1liwfxdx 1 0 where x y u OP is the standardized normal variable and t T u OP is the standardized truncation point in the N 01 scale Relationship between i and p If the selected character is normally distributed there is a simple relationship between i and p 1 1 2 Z tmexp7t t 1 2 pl q3tl Lomexp x dx Z in See the appendix of this chapter for the numerical values of the relationship among i tand p 164 When males and females have different selection differentials their average should be used S S selection d1fferent1al for males lead1ng to im 039 m P S S f selection differential for females leading to if f 039 P S Sm Sf leading to 139 im z39f Therefore R hZS h2Sm h2Sf hzimcp h2ifcp An example Given 6p 5 u 25 and p 020 find the truncation point selection intensity selection differential and response to selection 1 From the appendix table we get I 084 and z39 140 whenp 020 Note that tis the truncation point in the standardized scale What is the truncation point in the real scale T Because I THwe have Tt6Pu0842X525292l 6 F 2 Selection differential S 1390 l40gtlt5070 3 Assume that both the males and the females have the same selection intensity and h2 050 then the predicted selection response is R h2 S 05gtlt70 35 The population mean ofthe offspring will be u R 25 35 285 165 Truncation selection an AD YNuio 9 n1 5 gm gum Em a DUI p y u T Phenotypic value in the original scale y m m g xN g ml 01 Emma E 1 t gm p lt1gt Em a 76 4 72 u 2 a a XY 39 MVG Phenotypic value in the standardized normal scale x Chapter 3 Population Mean Phenotypic value the value observed when a character is measured on an individual For example the body weight of a particular pig is 300 lb The value of 300 1b is the phenotypic value of this pig for body weight Genotypic value average of the phenotypic values of all individuals who have the same genotype For example there are 1000 individuals in a population who have the same genotype A1142 For a particular trait the phenotypic values of the 1000 individuals are 125 089 210 Because all the 1000 individuals have the same genotype 1000 the average values of their phenotypic values 025 089 210 is the genotypic value of genotype A1142 Environmental deviation the difference between the phenotypic value and the genotypic value The rst genetic model P G E phenotypicwlue genotypicvalue environmentaldeviation Because environmental deviations can be positive and negative the average of E s for all individuals will be zero and thus the average phenotypic values is the genotypic value For the purpose of deduction we must assign arbitrary values to the genotypes under discussion This is done in the following way Considering a single locus with two alleles A1 and A2 we call the genotypic value of one homozygote a that of the other homozygote 7a and that of the heterozygote d We shall adopt the convention that11 is the allele that increases the value We thus have a scale of genotypic values as in Fig 31 Genotype A2142 141142 A1f1 Genotypic a 0 d a value Fig 31 Arbitrarily assigned genotypic values The genotypic value of the heterozygote al indicates the dominance effect If there is no dominance effect the genotypic value of A 1A2 should be half way between A1141 and 142142 ie al 0 With dominance d at 0 and the degree of dominance is usually expressed by the ratio d a Note the value 0 re ects the midpoint in the arbitrary scale The actual midpoint value can be different from 0 as shown in the following example The example also shows how to calculate the arbitrarily scaled genotypic values from the actual values Example dwar ng gene in the mouse Genotype A1A1 pgA1Az pg pgAzAz Actual weight g 14 12 6 14 6 10 The From this table we can calculate the actual midpoint value which is arbitrarily scaled genotypic values are then obtained by subtracting the actual midpoint value from the actual values of the genotypes After the subtractions we have al4 104 a6 10 4 d12 102 The degree of dominance is 6161 24 05 Note that midpoint is just a convenient way to express the position of the population There are several other ways to express the population position Remember to adjust the data by the midpoint before conducting any data analysis Population mean Let p and q be the frequencies of allele A1 and A2 respectively The population mean expectation is calculated using the following table Genotype Frequency y Value y y X f y 141A1 p2 a ap2 A1141 2m d dam AM q2 a aq2 The population mean in the arbitrary scale is the expectation of Y M EY 0P2 dam aq2 6100 q 2106161 The population mean I has been expressed as deviation from the actual midpoint Thus the actual population mean should be M midpoint Example dwar ng gene in the mouse Assume that the frequency of allele is p 09 and that ofpg allele is q 01 We have Map q2pqd4X09 0l2X09X0lx2356 The actual population mean isM midpoint 356 10 1356 Average effect of gene Genotypes are not transmitted from generation to generation it is the gene that is inherited from parent to offspring Average effect of a gene is the mean deviation from the population mean of individuals which received that gene from one parent the gene received from the other parent having come at random from the population First let us look at the average effect of allele A1 Consider one allele beingll and the other allele being randomly sampled from the population The probability that the other allele is A1 equals p If this happens the genotype will be A1141 with a genotypic value a On the other hand the probability that the other allele is A2 equals a Ifthe other allele is sampled then the genotype will be A1142 with a genotypic value 51 So in the population containing individuals with one A1 allele being fixed a proportion of p will be A1141 and a proportion ofq will be A1142 Define the mean ofthis population by M Al we have M A1 ap dq We then express this mean by the deviation from the population mean M resulting in the defined average e ect of genell all MA1Mqadqp The average ef ect of A2 is similarly defined by first obtaining M A pd qa and then 0 2 MA M padqp Average effect of gene substitution In a biallelic system the average effect of gene substitution is more informative than the average effect of gene The average effect of gene substitution is defined by a a1 a2 adq p The average gene effects and average effect of gene substitution can be derived using the conditional distribution and conditional mean approach Let us define a 2X2 table as Amp AM Amp A1A1apz A1A2d pq a A2Q 14214103 7 14214200 92 0 2 0L1 0L2 M A1 2 ap d apdq P P 001MAfM apdqap q2pqd qadqp Appendix E Linear Combination Quadratic Forms and Covariance Matrix Linear combination Let a be an ngtltl vector of constants Xan ngtltl vector of variables The scalar l aTX is called a linear combination ofX For example x1 13x1 4x2 5x3 3 4 5 x2 aTX x3 3 x1 where a 4 andX x2 5 x3 Quadratic forms LetA be an 71er constant matrix usually symmetrical X be an ngtltl vector of variables then Q X TAX is a quadratic form of X For example let x1 an a12 a13 x x2 A 6121 6122 6123 a 31 a 32 a 3 33 x1 all all 6113 X x2 and A all 6122 6123 x3 6131 6132 6133 the quadratic form is an all an 1 3 3 QXTAXC1 x2 x3 6121 6122 6123 2 ZZaUx1xj 11 1 a3 l X X X 3 Variance covariance matrix Let X x1 xn T be an ngtltl vector of multiple variables Denote the variance 2 and covariance by 61 Varxland GU C0vxlxy respectively for 1 l n The variancecovariance matrix of X is de ned as El 2 612 62 39 39 39 6m x VarX Var xn 6m 6H2 39 39 39 n VarX is a compact way of writing an array of variances and covariances Let Xx1xnT a vector 71 variables and Y y1ymT be a vector ofm variables De ne COVx1y1 C0Vx1y2 COVx1 ym Covx Covx Covx m CMXJT zyl 2y2 2y COVOCWJl OVOCWJz COVOCWJm W as the covariance matrix betweenX and Y We can de ne the following partitioned matrix X VarX E CovX Y T Var J E Y CovYXT VarY where VarX is an n X n variance matrix VarY is an m X m variance matrix and CovYXT CovXYT T is an m X n covariance matrix Expectation of linear combination Let Xx1xnT and Ex1 H1 fori ln De ne u1 HAT then x1 190 H1 EXE x E99 2 p 96 Exn H If a a1 any is a vector of constants then EaTX aTEX Let Y y1 ynT be another vector of variables we have EXYEXEY An example E2
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