lecture 2: basic population and quantitative genetics

Lecture 2: Basic Population and Quantitative Genetics

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

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

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

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

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)

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

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

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

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 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,

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

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

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

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?

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,

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,

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)

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]

Additive variance, VA, with no dominance (k = 0)

Allele frequency, p

VA

Complete dominance (k = 1)

Allele frequency, p

VA

VD

Overdominance (k = 2)

Allele frequency, p

Allele frequency, p

VAVD

Zero additivevariance

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