lecture 2: basic population and quantitative genetics
TRANSCRIPT
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Lecture 2: Basic Population and Quantitative Genetics
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Allele and Genotype Frequenciespi=freq(Ai)=freq(AiAi)+12Xi6=jfreq(AiAj)Given genotype frequencies, we can always compute allelefrequencies, e.g.,
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,freq(AiAj)=Ωp2ifori=j2pipjfori6=jHardy-Weinberg proportions
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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 randommating gives genotype frequencies in Hardy-Weinbergproportions, and they remain forever in these proportions
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Gametes and Gamete Frequenciesfreq(AABB)=freq(ABjfather)freq(ABjmother)freq(AaBB)=freq(ABjfather)freq(aBjmother)+freq(aBjfather)freq(ABjmother)When we consider two (or more) loci, we follow gametes
Under random mating, gametes combine at random, e.g.
Major complication: Even under HW conditions, gametefrequencies can change over time
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AB
AB
ab
ab
abAB
AB
abIn the F1, 50% AB gametes50 % 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
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Linkage disequilibriumfreq(AB)=freq(A)freq(B)freq(ABC)=freq(A)freq(B)freq(C)Random mating and recombination eventually changesgamete frequencies so that they are in linkage equilibrium (LE).once in LE, gamete frequencies do not change (unless actedon by other forces)
At LE, alleles in gametes are independent of each other:
When linkage disequilibrium (LD) present, alleles are nolonger independent --- knowing that one allele is in the gamete provides information on alleles at other locifreq(AB)6=freq(A)freq(B)The disequilibrium between alleles A and B is given byDAB=freq(AB)°freq(A)freq(B)
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freq(AB)=freq(A)freq(B)+DABD(t)=D(0)(1c)t°The Decay of Linkage Disequilibrium
The frequency of the AB gamete is given by
LE valueDeparture from LEIf recombination frequency between the A and B loci
is c, the disequilibrium in generation t is
Initial LD valueNote that D(t) -> zero, although the approach can beslow when c is very small
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Contribution of a locus to a trait
Basic model: P = G + E
Phenotypic value -- we will occasionallyalso use z for this value
Genotypic valueEnvironmental value
G = average phenotypic value for that genotypeif we are able to replicate it over the universeof environmental values
G - E covariance -- higher performing animalsmay be disproportionately rewarded
G x E interaction --- G values are differentacross environments. Basic model nowbecomes P = G + E + GE
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Alternative parameterizations of Genotypic values
Q1Q1 Q2Q1 Q2Q2
C C + a(1+k) C + 2aC C + a + d C + 2aC -a C + d C + a
2a = G(Q2Q2) - G(Q1Q1) d = ak =G(Q1Q2 ) - [G(Q2Q2) + G(Q1Q1) ]/2 d measures dominance, with d = 0 if the heterozygoteis exactly intermediate to the two homozygotes
k = d/a is a scaled measure of the dominance
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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
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Fisher’s Decomposition of GGij=πG+Æi+Æj+±ijOne of Fisher’s key insights was that the genotypic valueconsists of a fraction that can be passed from parent tooffspring and a fraction that cannot.πG=XGij¢freq(QiQj)Mean value, withAverage contribution to genotypic value for allele iSince parents pass along single alleles to their
offspring, the i (the average effect of allele i)represent these contributionsbGij=πG+Æi+ÆjThe genotypic value predicted from the individual
allelic effects is thusGij°Gij=±ijbDominance deviations --- the difference (for genotypeAiAj) between the genotypic value predicted from thetwo single alleles and the actual genotypic value,
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Gij=πG+2Æ1+(Æ2°Æ1)N+±ij2Æ1+(Æ2°Æ1)N=8><>:2Æ1forN=0;e.g,Q1Q1Æ1+Æ1forN=1;e.g,Q1Q22Æ1forN=2;e.g,Q2Q2Gij=πG+Æi+Æj+±ijFisher’s decomposition is a Regression
Predicted valueResidual errorA notational change clearly shows this is a regression,
Independent (predictor) variable N = # of Q2 allelesRegression slopeIntercept Regression residual
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0 1 2
N
G G22
G11
G21
Allele Q1 common, 2 > 1
Slope = 2 - 1
Allele Q2 common, 1 > 2Both Q1 and Q2 frequent, 1 = 2 = 0
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Genotype Q1Q1 Q2Q1 Q2Q2
Genotypicvalue
0 a(1+k) 2a
Consider a diallelic locus, where p1 = freq(Q1)πG=2p2a(1+p1k)Mean
Allelic effectsÆ2=p1a[1+k(p1°p2)]Æ1=°p2a[1+k(p1°p2)]Dominance deviations±ij=Gij°πG°Æi°Æj
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Average effects and Breeding ValuesBV(Gij)=Æi+ÆjBV=nXk=1≥Æ(k)i+Æ(k)k¥( )
The values are the average effects of an allele
Breeders focus on breeding value (BV)
Why all the fuss over the BV?
Consider the offspring of a QxQy sire matedto a random dam. What is the expected valueof the offspring?
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GenotypeFrequencyValueQxQw1/4πG+Æx+Æw+±xwQxQz1/4πG+Æx+Æz+±xzQyQw1/4πG+Æy+Æw+±ywQyQz1/4πG+Æy+Æz+±yzπO=πG+µÆx+Æy2∂+µÆw+Æz2∂+±xw+±xz+±yw+±yz4The expected value of an offspring is the expected value of
For a random dam, these have expected value 0For random w and z alleles, this has an expected value of zeroπO°πG=µÆx+Æy2∂=BV(Sire)2Hence,
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BV(Sire)=2(π0°πG)π0°πG=BV(Sire)2+BV(Dam)2We can thus estimate the BV for a sire by twicethe deviation of his offspring from the pop mean,
More generally, the expected value of an offspringis the average breeding value of its parents,
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Genetic VariancesGij=πg+(Æi+Æj)+±ijæ2(G)=nXk=1æ2(Æ(k)i+Æ(k)j)+nXk=1æ2(±(k)ij)æ2G=æ2A+æ2Dæ2(G)=æ2(πg+(Æi+Æj)+±ij)=æ2(Æi+Æj)+æ2(±ij)As Cov() = 0
Additive Genetic Variance(or simply Additive Variance)
Dominance Genetic Variance(or simply dominance variance)
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æ2D=2E[±2]=mXi=1mXj=1±2ijpipjæ2D=(2p1p2ak)2æ2A=2p1p2a2[1+k(p1°p2)]2One locus, 2 alleles:
One locus, 2 alleles:
Q1Q1 Q1Q2 Q2Q2
0 a(1+k) 2a
Dominance effects additive variance
When dominance present, asymmetric function of allele Frequencies
Equals zero if k = 0This is a symmetric function ofallele frequencies
æ2A=2E[Æ2]=2mXi=1Æ2ipiSince E[] = 0, Var() = E[( -a)2] = E[2]
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Additive variance, VA, with no dominance (k = 0)
Allele frequency, p
VA
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Complete dominance (k = 1)
Allele frequency, p
VA
VD
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Overdominance (k = 2)
Allele frequency, p
Allele frequency, p
VAVD
Zero additivevariance
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EpistasisGijkl=πG+(Æi+Æj+Æk+Æl)+(±ij+±kj)+(ÆÆik+ÆÆil+ÆÆjk+ÆÆjl)+(Ʊikl+Ʊjkl+Ʊkij+Ʊlij)+(±±ijkl)=πG+A+D+AA+AD+DBreeding valueDominance value -- interaction
between the two alleles at a locusAdditive x Additive interactions --interactions between a single alleleat one locus with a single allele at another
Additive x Dominant interactions --interactions between an allele at onelocus with the genotype at another, e.g.allele Ai and genotype Bkj
Dominance x dominance interaction ---the interaction between the dominancedeviation at one locus with the dominancedeviation at another.
These components are defined to be uncorrelated,(or orthogonal), so thatæ2G=æ2A+æ2D+æ2AA+æ2AD+æ2DD