Introduction to GeneticAnalysis
Ecology and Evolutionary Biology,University of Arizona
Adjunct Appointments Molecular and Cellular Biology
Plant SciencesEpidemiology & Biostatistics
Animal Sciences
Bruce Walsh
Outline• Mendelian Genetics
– Genes, Chromosomes & DNA– Mendel’s laws– Linkage– Linkage disequilibrium
• Quantitative Genetics– Fisher’s decomposition of Genetic value– Fisher decomposition of Genetic Variances– Resemblance between relatives– Searching for the underlying genes
Mendelian Genetics
Following a single (or several) genes that we can directly
score
Phenotype highly informative as to genotype
Mendel’s GenesGenes are discrete particles, with each parent passingone copy to its offspring.
Let an allele be a particular copy of a gene. In Diploids,each parent carries two alleles for every gene, onefrom each parent
Each parent contributes one of its two alleles (atrandom) to its offspring
For example, a parent with genotype Aa (a heterozygote for alleles A and a) has a 50% probability of passing anA allele onto its offspring and a 50% probability ofpassing along an a allele.
Example: Pea seed color
Mendel found that his pea lines differed in seed color,with a single locus (with alleles Y and g) determining green vs. yellow
YY (Y homozygote) --> yellow phenotypeYg (heterozygote) --> yellow phenotypegg (g homozygote) --> green phenotype
Note that in this simple case, each genotype mapsto a single phenotype
Likewise, the phenotype can tell us about the underlyingGenotype. Green = gg, Yellow = carries Y allele
Y is dominant to g, g is recessive to Y
Cross Yg x Yg. Offspring are 1/4 YY, 1/2 Yg, 1/4 gg3/4 yellow peas, 1/4 green peas
Cross Yg x gg. Offspring are 1/2Yg, 1/2 gg,1/2 yellow, 1/2 green
Dealing with two (or more) genes
For 7 pea traits, Mendel observed Independent Assortment
The genotype at one locus is independent of the second
RR, Rr - round seeds, rr - wrinkled seeds
Pure round, green (RRgg) x pure wrinkled yellow (rrYY)
F1 --> RrYg = all round, yellow
What about the F2?
YY, Yg - yellow seeds, gg - green seeds
Let R- denote RR and Rr. R- are round. Note in F2,Pr(R-) = 1/2 + 1/4 = 3/4, Pr(rr) = 1/4
Likewise, Y- are YY or Yg, and are yellow
Phenotype Genotype Frequency
Yellow, round Y-R- (3/4)*(3/4) = 9/16
Yellow, wrinkled Y-rr (3/4)*(1/4) = 3/16
Green, round ggR- (1/4)*(3/4) = 3/16
Green, wrinkled ggrr (1/4)*(1/4) = 1/16
Or a 9:3:3:1 ratio
Mendel was wrong: Linkage
Phenotype
Genotype Observed Expected
Purple long P-L- 284 215
Purple round
P-ll 21 71
Red long ppL- 21 71
Red round ppll 55 24
Bateson and Punnet looked at flower color: P (purple) dominant over p (red )
pollen shape: L (long) dominant over l (round)
Excess of PL, pl gametes over Pl, pL
Departure from independent assortment
PPLL x ppll --> PL/pl F1
Chromosomal theory of inheritance
It was soon postulated that Genes are carried on chromosomes, because chromosomes behaved in afashion that would generate Mendel’s laws.
Early light microscope work on dividing cells revealedsmall (usually) rod-shaped structures that appear topair during cell division. These are chromosomes.
We now know that each chromosome consists of asingle double-stranded DNA molecule (covered withproteins), and it is this DNA that codes for the genes.
Humans have 23 pairs of chromosomes (for a total of 46)
22 pairs of autosomes (chromosomes 1 to 22)1 pair of sex chromosomes -- XX in females, XY in males
Humans also have another type of DNA molecule, namelythe mitochondrial DNA genome that exists in tens to thousands of copies in the mitochondria present in all ourcells
mtDNA is usual in that it is strictly maternally inherited.Offspring get only their mother’s mtDNA.
Linkage
If genes are located on different chromosomes they(with very few exceptions) show independent assortment.
Indeed, peas have only 7 chromosomes, so was Mendel luckyin choosing seven traits at random that happen to allbe on different chromosomes? Problem: compute this probability.
However, genes on the same chromosome, especially ifthey are close to each other, tend to be passed ontotheir offspring in the same configuation as on theparental chromosomes.
Consider the Bateson-Punnet pea data
Let PL / pl denote that in the parent, one chromosomecarries the P and L alleles (at the flower color andpollen shape loci, respectively), while the other chromosome carries the p and l alleles.
Unless there is a recombination event, one of the twoparental chromosome types (PL or pl) are passed ontothe offspring. These are called the parental gametes.
However, if a recombination event occurs, a PL/pl parent can generate Pl and pL recombinant gametesto pass onto its offspring.
Linkage --> excess of parental gametes
Let c denote the recombination frequency --- theprobability that a randomly-chosen gamete from theparent is of the recombinant type (i.e., it is not aparental gamete).
For a PL/pl parent, the gamete frequencies are
Gamete type Frequency Expectation under independent assortment
PL (1-c)/2 1/4
pl (1-c)/2 1/4
pL c/2 1/4
Pl c/2 1/4
Parental gametes in excess, as (1-c)/2 > 1/4 for c < 1/2Recombinant gametes in deficiency, as c/2 < 1/4 for c < 1/2In Bateson data, Freq(ppll) = 55/381 =0.144. Freq(ppll) = [(1-c)/2]2,Solving gives c = 0.24
Linkage is our friendWhile linkage (at first blush) may seem a complication, itis actually our friend, allowing us to map genes --- determining which genes are on which chromosomes and also fine-mapping their position on a particular chromosome
Historically, the genes that have been mapped havedirect effects on phenotypes (pea color, fly eye color,any number of simple human diseases, etc. )
In the molecular era, we are often concerned withmolecular markers, variations in the DNA sequence thattypically have no effect on phenotype
Molecular MarkersYou and your neighbor differ at roughly 22,000,000 nucleotides (base pairs) out of the roughly 3 billionbp that comprises the human genome
Hence, LOTS of molecular variation to exploit
SNP -- single nucleotide polymorphism. A particularposition on the DNA (say base 123,321 on chromosome 1)that has two different nucleotides (say G or A) segregating
STR -- simple tandem arrays. An STR locus consists ofa number of short repeats, with alleles defined bythe number of repeats. For example, you might have6 and 4 copies of the repeat on your two chromsome 2s
Gametes and Gamete Frequencies
freq(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.
Linkage disequilibrium
freq(AB) = freq(A) freq(B)freq(ABC) = freq(A)freq(B)freq(C)
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 loci
freq(AB)6= freq(A) freq(B)
The disequilibrium between alleles A and B is given by
DA B = freq(AB) ° freq(A)freq(B)
Forces that Generate LD
• Selection• Drift• Migration (admixture)• Mutation• Population structure (stratification)
freq(AB) = freq(A) freq(B) + DAB
D(t) = D(0)(1 c)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
Not surprising that very tightly-linked markers (c <<0.01) are often in LD
Key Mendelian Concepts• Genes, Chromosomes & DNA• “Classical” vs Molecular markers• Linkage
– Parental gametes in excess. Alleles at nearby loci tend to segregate together
• Linkage disequilibrium (LD)– Excess of parental gametes seen in any particular
cross– LD implies in the population that there is a non-
random association of allele– Unlinked alleles can show LD due to population
structure
Quantitative Genetics
The analysis of traits whose variation is determined by
both a number of genes and environmental factors
Phenotype is highly uninformative as tounderlying genotype
Complex (or Quantitative) trait
• No (apparent) simple Mendelian basis for variation in the trait
• May be a single gene strongly influenced by environmental factors
• May be the result of a number of genes of equal (or differing) effect
• Most likely, a combination of both multiple genes and environmental factors.
• Example: Blood pressure, cholesterol levels– Known genetic and environmental risk factors
Phenotypic distribution of a traitConsider a specific locus influencing the trait
For this locus, mean phenotype = 0.15, whileoverall mean phenotype = 0
Goals of Quantitative Genetics
• Partition total trait variation into genetic (nature) vs. environmental (nurture) components
• Predict resemblance between relatives– If a sib has a disease/trait, what are your odds?
• Find the underlying loci contributing to genetic variation – QTL -- quantitative trait loci
• Deduce molecular basis for genetic trait variation
Dichotomous (binary) traits
Presence/absence traits (such as a disease) can still(and usually do) have a complex genetic basis
Consider a DS locus underlying a disease, with allelesD and d, where allele D significantly increases yourdisease risk
In particular, Pr(disease | DD) = 0.5, so that thePenetrance of genotype DD is 50%
Suppose Pr(disease | Dd ) = 0.2, Pr(disease | dd) = 0.05
dd individuals can rarely display the disease, largelybecause of exposure to adverse environmental conditions
If freq(d) = 0.9, what is Prob (DD | show disease) ?
freq(disease) = 0.12*0.5 + 2*0.1*0.9*0.2 + 0.92*0.05 = 0.0815
From Bayes’ theorem, Pr(DD | disease) = Pr(disease |DD)*Pr(DD)/Prob(disease) = 0.12*0.5 / 0.0815 = 0.06 (6 %)
dd individuals can give rise to phenocopies 5% of the time,showing the disease but not as a result of carrying therisk allele
Pr(Dd | disease) = 0.442, Pr(dd | disease) = 0.497
Basic model of Quantitative Genetics
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[P]
G x E interaction --- G values are differentacross environments. Basic model nowbecomes P = G + E + GE
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
Contribution of a locus to a trait
Example: Apolipoprotein E & Alzheimer’s
Genotype ee Ee EE
Average age of onset
68.4 75.5 84.3
2a = G(EE) - G(ee) = 84.3 - 68.4 --> a = 7.95
ak =d = G(Ee) - [ G(EE)+G(ee)]/2 = -0.85
k = d/a = 0.10 Only small amount of dominance
Fisher’s (1918) Decomposition of G
One 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 =X
Gi j ¢freq(QiQj )Mean value, withAverage contribution to genotypic value for allele iSince parents pass along single alleles to theiroffspring, the i (the average effect of allele i)represent these contributions
Gi j = πG +Æi +Æj +±i j
bGi j = πG +Æi +Æj
The genotypic value predicted from the individualallelic effects is thus
G i j ° Gi j =±i jb
Dominance deviations --- the difference (for genotypeAiAj) between the genotypic value predicted from thetwo single alleles and the actual genotypic value,
Consider the genotypic value Gij resulting from an AiAj individual
Gi j = πG +2Æ1 + (Æ2 ° Æ1)N +±i j
2Æ1 + (Æ2 ° Æ1)N =
8><
>:
2Æ1 forN =0; e.g, Q1Q1
Æ1 +Æ1 forN =1; e.g, Q1Q2
2Æ1 forN =2; e.g, Q2Q2
Gi j = πG +Æi +Æj +±i j
Fisher’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 effects
Æ2 = p1a[1+k (p1 ° p2 ) ]
Æ1 = °p2a[1+ k (p1 ° p2 )]Dominance deviations±i j = G i j ° πG ° Æi ° Æj
Average effects and Additive Genetic Values
A (Gi j ) =Æi +ÆjA =nX
k=1
≥Æ(k)
i +Æ(k)k
¥( )
The values are the average effects of an allele
A key concept is the Additive Genetic Value (A) ofan individual
Why all the fuss over A?
Suppose father has A = 10 and mother has A = -2for (say) blood pressure
Expected blood pressure in their offspring is (10-2)/2 = 4 units above the population mean. Offspring A =Average of parental A’s
KEY: parents only pass single alleles to their offspring.Hence, they only pass along the A part of their genotypicValue G
Genetic Variances
Gi j = πg + (Æi +Æj ) +±i j
æ2(G) =nX
k=1
æ2(Æ(k)i +Æ(k)
j ) +nX
k=1
æ2(±(k)i j )
æ2G =æ2
A +æ2D
æ2(G) =æ2(πg +(Æi +Æj ) +±i j ) =æ2(Æi +Æj ) +æ2(±i j)
As Cov() = 0
Additive Genetic Variance(or simply Additive Variance)
Dominance Genetic Variance(or simply dominance variance)
Key concepts (so far)• i = average effect of allele i
– Property of a single allele in a particular population (depends on genetic background)
• A = Additive Genetic Value (A) – A = sum (over all loci) of average effects– Fraction of G that parents pass along to their offspring– Property of an Individual in a particular population
• Var(A) = additive genetic variance– Variance in additive genetic values– Property of a population
• Can estimate A or Var(A) without knowing any of the underlying genetical detail (forthcoming)
æ2D = 2E[±2] =
mX
i=1
mX
j=1
±2i j pi pj
æ2D = (2p1p2 ak)2
æ2A = 2p1p2 a2[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 ] = 2
mX
i=1
Æ2i pi
Since E[] = 0, Var() = E[( -a)2] = E[2]
Epistasis
Gi j kl = πG + (Æi +Æj +Æk +Æl) + (±i j +±k j )
+ (ÆÆik +ÆÆi l +ÆÆjk +ÆÆj l)+ (Ʊikl +Ʊjkl +Ʊki j +Ʊl i j )+ (±±i j kl)
= πG + A + D + AA + AD + DD
Additive Genetic valueDominance value -- interactionbetween the two alleles at a locus
Additive 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 that
æ2G =æ2
A +æ2D +æ2
AA +æ2AD +æ2
D D
Heritability• Central concept in quantitative genetics• Proportion of variation due to additive genetic
values – h2 = VA/VP
– Phenotypes (and hence VP) can be directly measured
– Breeding values (and hence VA ) must be estimated
• Estimates of VA require known collections of relatives
Key observations
• The amount of phenotypic resemblance among relatives for the trait provides an indication of the amount of genetic variation for the trait.
• If trait variation has a significant genetic basis, the closer the relatives, the more similar their appearance
Genetic Covariance between relatives
Genetic covariances arise because two related individuals are more likely to share alleles than are two unrelated individuals.
Sharing alleles means having alleles that are identical by descent (IBD): both copies of can be traced back to a single copy in a recent common ancestor.
Father Mother
No alleles IBD One allele IBDBoth alleles IBD
Parent-offspring genetic covariance
Cov(Gp, Go) --- Parents and offspring share EXACTLY one allele IBD
Denote this common allele by A1
Gp = Ap + Dp =Æ1 +Æx + D1x
Go = Ao + Do =Æ1 +Æy + D1y
IBD alleleNon-IBD alleles
Cov(Go;Gp) = Cov(Æ1 +Æx + D1x;Æ1 +Æy + D1y
= Cov(Æ1;Æ1) + Cov(Æ1;Æy) +Cov(Æ1; D1y)+ Cov(Æx;Æ1) +Cov(Æx;Æy) +Cov(Æx; D1y)
+ Cov(D1x;Æ1) + Cov(D1x;Æy) + Cov(D1x; D1y)
All white covariance terms are zero.
• By construction, and D are uncorrelated
• By construction, from non-IBD alleles are uncorrelated
• By construction, D values are uncorrelated unless both alleles are IBD
Cov(Æx;Æy) =Ω0 if x6=y; i.e., not IBD
Var(A)=2 if x =y; i.e., IBD
Var(A) = Var(Æ1 +Æ2) = 2Var(Æ1)
so thatVar(Æ1) = Cov(Æ1;Æ1) = Var(A)=2
Hence, relatives sharing one allele IBD have agenetic covariance of Var(A)/2
The resulting parent-offspring genetic covariance becomes Cov(Gp,Go) = Var(A)/2
Half-sibs
The half-sibs share one allele IBD • occurs with probability 1/2
1
o1
2
o2
The half-sibs share no alleles IBD • occurs with probability 1/2
Each sib gets exactly one allele from common father,different alleles from the different mothers
Hence, the genetic covariance of half-sibs is just (1/2)Var(A)/2 = Var(A)/4
Full-sibsFather Mother
Full SibsPaternal allele not IBD [ Prob = 1/2 ]Maternal allele not IBD [ Prob = 1/2 ]-> Prob(zero alleles IBD) = 1/2*1/2 = 1/4
Paternal allele IBD [ Prob = 1/2 ]Maternal allele IBD [ Prob = 1/2 ]-> Prob(both alleles IBD) = 1/2*1/2 = 1/4
Prob(exactly one allele IBD) = 1/2= 1- Prob(0 IBD) - Prob(2 IBD)
Each sib getsexact one allelefrom each parent
IB D alleles Probability Contr ibution
0 1/ 4 0
1 1/ 2 Var(A)/ 2
2 1/ 4 Var(A) + Var(D)
IBD alleles Probability Contribution
0 1/4 0
1 1/2 Var(A)/2
2 1/4 Var(A) + Var(D)
Resulting Genetic Covariance between full-sibs
Cov(Full-sibs) = Var(A)/2 + Var(D)/4
Genetic Covariances for General Relatives
Let r = (1/2)Prob(1 allele IBD) + Prob(2 alleles IBD)
Let u = Prob(both alleles IBD)
General genetic covariance between relativesCov(G) = rVar(A) + uVar(D)
When epistasis is present, additional terms appearr2Var(AA) + ruVar(AD) + u2Var(DD) + r3Var(AAA) +
Sample Covariances
Cov(monozygotic twins) = VA + VD + Cov(E)
Cov(dizygotic twins) = VA/2 + VD/4 + Cov(E)
Cov(parent, offspring) = VA/2
Hence, can estimate genetic variance componentsFrom phenotypic covariances using known sets of relatives
More generally, use all comparisons between relatives ina complex pedigree (REML estimate of variances)
Relative risks for binary traitsLet z1 and z2 denote the trait state (0,1) in tworelatives.
Recurrence risk, KR (for relatives of type R) = Prob(z2 =1 | z1 = 1)
James’ identity: KR = K + Cov(z1,z2)/K where K = Prob(z=1), i.e., the population prevalence
Relative risk, R = KR/K
Risch’s identity: R = 1 + Cov(z1,z2)/K2
Searching for QTLs: Marker-Trait Associations
Key: With linkage = excess of parential gametes
MQ/mq father -- M associated with QTL alleleQ (which increases trait value over q). Comparingmean trait values in offspring for paternal-M vs. paternal-m will show (for sufficiently large sample) a significant difference.
Since the phase may differ across parents (e.g.,mother might be Mq/mQ), critical to contrast markers alleles from each parent separately
I. Within a pedigree
Searching for QTLs: Marker-Trait Associations
II. Population-level linkage disequilibrium
Key: With LD, covariance between alleles
For very tightly-linked markers (less than 1 cM), mightexpect some population-level disequilibrium
Hence, can contrast (say) M vs. m grouped over allindividuals to look for a difference in trait value btwthe two groups.
If marker locus is sufficiently close to a QTL, LD mightbe present and an marker-trait association detected.
Complication: Population structure can generate acovariance btw unlinked markers
Key concepts• P = G + E = A + D + I + E• Var(G) = Var(A) + Var(D) + Var(I)• Phenotypic covariances can be used to
estimate components of Var(G)• h2 = Var(A)/Var(P) is the heritability of a
trait, measure of how parents & offspring resemble each other
• Can use linkage (within a pedigree) or linkage disequilibrium (within a population) to search for QTLs via marker-trait associations