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Lecture 2: Basic Population and Quantitative Genetics Allele and Genotype Frequencies Given genotype frequencies, we can always compute allele frequencies, e.g., 1X p i = freq(A i ) = freq(Ai Ai ) + freq(Ai A j ) 2 i6 j = The converse is not true: given allele frequencies we cannot uniquely determine the genotype frequencies For n alleles, there are n(n+1)/2 genotypes If we are willing to assume random mating, ž p2 i for i = j Hardy-Weinberg freq(Ai Aj ) = proportions 2pi pj = for i 6 j Hardy-Weinberg • Prediction of genotype frequencies from allele freqs • Allele frequencies remain unchanged over generations, provided: • Infinite population size (no genetic drift) • No mutation • No selection • No migration • Under HW conditions, a single generation of random mating gives genotype frequencies in Hardy-Weinberg proportions, and they remain forever in these proportions Gametes and Gamete Frequencies When we consider two (or more) loci, we follow gametes Under random mating, gametes combine at random, e.g. freq(AAB B ) = freq(AB jfat her) freq(AB jmot her) freq(AaB B ) = freq(AB jfat her) freq(aB jmot her) + freq(aB jfat her) freq(AB jmot her) Major complication: Even under HW conditions, gamete frequencies can change over time AB ab AB ab AB ab AB In the F1, 50% AB gametes ab 50 % ab gametes If A and B are unlinked, the F2 gamete frequencies are AB 25% ab 25% Ab 25% aB 25% Thus, even under HW conditions, gamete frequencies change Linkage disequilibrium Random mating and recombination eventually changes gamete frequencies so that they are in linkage equilibrium (LE). once in LE, gamete frequencies do not change (unless acted on by other forces) At LE, alleles in gametes are independent of each other: When linkage disequilibrium (LD) present, alleles are no freq(AB --- freq(A) freq(B ) longer independent ) =knowing that one allele is in the freq(ABC) = freq(A) freq(B) freq(C) gamete provides information on alleles at other loci The disequilibrium between alleles A and B is given by = freq(AB ) 6 freq(A) freq(B ) D A B = freq(AB ) ° freq(A) freq(B ) The Decay of Linkage Disequilibrium The frequency of the AB gamete is given by freq(AB ) = freq(A) freq(B ) + D A B Departure from A If recombination frequency between theLE and B loci LE value is c, the disequilibrium in generation t is D (t ) = D (0)(1 ° c) t Initial although Note that D(t) -> zero, LD value the approach can be slow when c is very small Contribution of a locus to a trait Basic model: P = G + E Genotypic -- we value Phenotypic valuevalue will occasionally Environmental also use z value for that G = average phenotypic for this value genotype if we are able to replicate it over the universe of environmental values G - E covariance -- higher performing animals may be disproportionately rewarded G x E interaction --- G values are different across environments. Basic model now becomes P = G + E + GE Alternative parameterizations of Genotypic values Q1Q1 Q2Q1 Q2Q2 C C + a(1+k) C + 2a C C+a+d C + 2a C -a C+d C+a d = ak =G(Q1Q2 G(Q2 with d ) = 1Q1 1Q1 ]/2 d measures dominance, Q )2QG(Q0 if) the)heterozygote 2a = ) - [G(Q - 2 + G(Q 2 is exactly intermediate to the two homozygotes k = d/a is a scaled measure of the dominance Example: Booroola (B) gene Genotype bb Bb BB Average Litter size 1.48 2.17 2.66 2a = G(BB) - G(bb) = 2.66 -1.46 --> a = 0.59 ak =d = G(Bb) - [ G(BB)+G(bb)]/2 = 0.10 k = d/a = 0.17 Fisher’s Decomposition of G One of Fisher’s key insights was that the genotypic value consists of a fraction that can be passed from parent to offspring and a fraction that cannot. G i j = š G + Æ + Æ + ±i j i j X Average contribution --- genotypic j valueto their The genotypic value along = difference (forallele Since parents with to the Mean value, pass š G singleG i the fori genotype Dominance deviationspredicted from¢freq(Q Q ) i alleles individual j allelic effectsthei genotypic value predicted from Aj) between is (the b Aioffspring, the athus average effect of allele i) the two single alleles and the actualš G + Æ value,j G i j = genotypic + Æ represent these contributions i b G i j ° Gi j = ±i j Fisher’s decomposition is a Regression G i j = š G + Æ + Æ + ±i j i j A notational change clearly showsResidual error Predicted value this is a regression, G i j = š G + 2Æ + (Æ ° Æ )N + ±i j 1 2 1 Intercept 8 Regression = # Independent (predictor) variable Nslopeof Q2 alleles Regression residual > 2Æ < 1 forN = 0; e.g, Q 1 Q 1 2Æ + (Æ ° Æ )N = Æ + Æ 1 2 1 forN = 1; e.g, Q 1 Q 2 > 1 : 1 2Æ1 forN = 2; e.g, Q 2 Q 2 Both Q2 and Q2 a2 > a1 Allele Q11 common,frequent, a1 = a2 = 0 common, a1 2 G21 Slope = a2 - a1 G G22 G11 0 1 2 N Consider a diallelic locus, where p1 = freq(Q1) Genotype Q1Q1 Q2Q1 Q2Q2 Genotypic 0 a(1+k) 2a value Mean š G = 2p2 a(1 + p1 k) Allelic effects Æ = p1 a [ 1 + k ( p1 ° p2 ) ] 2 Æ = ° p2 a [1 + k ( p1 ° p2 ) ] 1 Dominance deviations ±i j = G i j ° š G ° Æ ° Æ i j Average effects and Breeding Values The a values are the average effects of an allele Breeders focus on breeding value (BV) X • (k) n ¥ B V(Gi j ) =ÆÆ++ (kÆ BV = i i k= 1 ( k) Æ j ) Why all the fuss over the BV? Consider the offspring of a QxQy sire mated to a random dam. What is the expected value of the offspring? Genot ype Frequency Value Q x Qw 1/ 4 šG + Æ + x Æ + ±x w w Q x Qz 1/ 4 šG + Æ + x Æ + ±x z z Qy Qw 1/ 4 šG + Æ + y Æ + ±y w w Qy Qz 1/ 4 šG + Æ + y Æ + ±y z z The expected value of an offspring is the expected value of µ Ž µ Ž Æ +Æ x y Æ + Æ w z ±x w + ±x z + ±yw + ±yz šO = šG + + + 2 2 4 For random these have expected value For a random dam, w and z alleles, this has an 0 expected value of zero Hence, µ Ž Æ + Æ x y B V(Sire) šO ° šG = = 2 2 We can thus estimate the BV for a sire by twice the deviation of his offspring from the pop mean, B V(Sire) = 2(š 0 ° š G ) More generally, the expected value of an offspring is the average breeding value of its parents, B V(Sire) B V(Dam) š0 ° šG = + 2 2 Genetic Variances G i j = š g + (Æ + Æ ) + ±i j i j æ2 (G) = æ2 (š g X (Æ + Æ ) + ±i j ) = æ2n i + Æ ) + æ2 (±i j ) n + i j X(Æ j ( k) (k) ( k) æ2 (G) = æ2 (Æ + Æ )+ æ2 (±i j ) i As Cov(a,d) = 0 j k= 1 k= 1 2 æG = 2 æA + 2 æD Dominance Genetic Variance Additive Genetic Variance (or simply dominance (or simply Additive Variance) variance) m X Q1Q1 Q1Q2 Q2Q2 æ2 = 2E [Æ ] = 2 2 2 Æ pi Since E[a] = 0, Var(a)0 E[(aa(1+k)] = E[a2a = -ma)2 2] A i i= 1 One locus, 2 alleles: 2 æA = 2p 1 p2 a2 [ 1 + k ( p 1 ° p2 ) ]2 When dominance present, Dominance effects m X Xm asymmetric function of allele additive variance 2 = 2E [±2 ] = æD ±i j i jFrequencies 2 p p i= 1 j= 1 One locus, 2 alleles: 2 æD = (2p 1 p2 ak)2 Equals zero if k = 0 This is a symmetric function of allele frequencies Additive variance, VA, with no dominance (k = 0) VA Allele frequency, p Complete dominance (k = 1) VA VD Allele frequency, p Overdominance (k = 2) VA VD Zero additive variance Allele frequency, p Allele frequency, p Epistasis G i j kl = š G + (Æ + Æ + Æ + Æ ) + (±i j + ±k j ) i j k l Æ Æ Æ Æ + (Æ i k + Æ i l + Æ j k + Æ j l ) ± ± ± ± + (Æ i k l + Æ j k l + Æ ki j + Æ l i j ) + (±±i j k l ) = š G + A + D + AA + AD + D These components are Additivetointeractions ----- Dominance xvalue interactions Additive x Dominant be uncorrelated, Dominancedominance interaction Breeding value Additive x defined -- interaction-- (or orthogonal), interaction between the dominance the so that between a single allele interactions between an allele locus between interactionsthe two alleles at a at one at onewith the genotype at the dominance locus locus with a single allele at another deviation at one locus with another, e.g. 2 allele2 i andanother. B2 + æD at genotype kj A æ2 = æA deviation+ æA A + æA D + æ2 D G 2 D