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Introduction to Genetic Analysis

Introduction to Genetic Analysis. Bruce Walsh. jbwalsh@u.arizona.edu. Ecology and Evolutionary Biology, University of Arizona. Adjunct Appointments Molecular and Cellular Biology Plant Sciences Epidemiology & Biostatistics Animal Sciences. Outline. Mendelian Genetics

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Introduction to Genetic Analysis

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  1. Introduction to GeneticAnalysis Bruce Walsh jbwalsh@u.arizona.edu Ecology and Evolutionary Biology, University of Arizona Adjunct Appointments Molecular and Cellular Biology Plant Sciences Epidemiology & Biostatistics Animal Sciences

  2. 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

  3. Mendelian Genetics Following a single (or several) genes that we can directly score Phenotype highly informative as to genotype

  4. Mendel’s Genes Genes are discrete particles, with each parent passing one copy to its offspring. Let an allele be a particular copy of a gene. In Diploids, each parent carries two alleles for every gene, one from each parent Each parent contributes one of its two alleles (at random) to its offspring For example, a parent with genotype Aa (a heterozygote for alleles A and a) has a 50% probability of passing an A allele onto its offspring and a 50% probability of passing along an a allele.

  5. 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 phenotype Yg (heterozygote) --> yellow phenotype gg (g homozygote) --> green phenotype Note that in this simple case, each genotype maps to a single phenotype Y is dominant to g, g is recessive to Y Likewise, the phenotype can tell us about the underlying Genotype. Green = gg, Yellow = carries Y allele Cross Yg x Yg. Offspring are 1/4 YY, 1/2 Yg, 1/4 gg 3/4 yellow peas, 1/4 green peas Cross Yg x gg. Offspring are 1/2Yg, 1/2 gg, 1/2 yellow, 1/2 green

  6. 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 YY, Yg - yellow seeds, gg - green seeds Pure round, green (RRgg) x pure wrinkled yellow (rrYY) F1 --> RrYg = all round, yellow What about the F2?

  7. 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 Or a 9:3:3:1 ratio

  8. Mendel was wrong: Linkage Bateson and Punnet looked at flower color: P (purple) dominant over p (red ) pollen shape: L (long) dominant over l (round) PPLL x ppll --> PL/pl F1 Excess of PL, pl gametes over Pl, pL Departure from independent assortment

  9. Chromosomal theory of inheritance Early light microscope work on dividing cells revealed small (usually) rod-shaped structures that appear to pair during cell division. These are chromosomes. It was soon postulated that Genes are carried on chromosomes, because chromosomes behaved in a fashion that would generate Mendel’s laws. We now know that each chromosome consists of a single double-stranded DNA molecule (covered with proteins), and it is this DNA that codes for the genes.

  10. 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, namely the mitochondrial DNA genome that exists in tens to thousands of copies in the mitochondria present in all our cells mtDNA is usual in that it is strictly maternally inherited. Offspring get only their mother’s mtDNA.

  11. 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 lucky in choosing seven traits at random that happen to all be on different chromosomes? Problem: compute this probability. However, genes on the same chromosome, especially if they are close to each other, tend to be passed onto their offspring in the same configuation as on the parental chromosomes.

  12. Consider the Bateson-Punnet pea data Let PL / pl denote that in the parent, one chromosome carries the P and L alleles (at the flower color and pollen shape loci, respectively), while the other chromosome carries the p and l alleles. Unless there is a recombination event, one of the two parental chromosome types (PL or pl) are passed onto the offspring. These are called the parental gametes. However, if a recombination event occurs, a PL/pl parent can generate Pl and pL recombinant gametes to pass onto its offspring. Linkage --> excess of parental gametes

  13. Recombinant gametes in deficiency, as c/2 < 1/4 for c < 1/2 Parental gametes in excess, as (1-c)/2 > 1/4 for c < 1/2 Let c denote the recombination frequency --- the probability that a randomly-chosen gamete from the parent is of the recombinant type (i.e., it is not a parental gamete). For a PL/pl parent, the gamete frequencies are In Bateson data, Freq(ppll) = 55/381 =0.144. Freq(ppll) = [(1-c)/2]2, Solving gives c = 0.24

  14. Linkage is our friend While linkage (at first blush) may seem a complication, it is 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 have direct effects on phenotypes (pea color, fly eye color, any number of simple human diseases, etc. ) In the molecular era, we are often concerned with molecular markers, variations in the DNA sequence that typically have no effect on phenotype

  15. Molecular Markers You and your neighbor differ at roughly 22,000,000 nucleotides (base pairs) out of the roughly 3 billion bp that comprises the human genome Hence, LOTS of molecular variation to exploit SNP -- single nucleotide polymorphism. A particular position 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 of a number of short repeats, with alleles defined by the number of repeats. For example, you might have 6 and 4 copies of the repeat on your two chromsome 2s

  16. f r e q ( A A B B ) = f r e q ( A B j f a t h e r ) f r e q ( A B j m o t h e r ) f r e q ( A a B B ) = f r e q ( A B j f a t h e r ) f r e q ( a B j m o t h e r ) + f r e q ( a B j f a t h e r ) f r e q ( A B j m o t h e r ) Gametes and Gamete Frequencies When we consider two (or more) loci, we follow gametes Under random mating, gametes combine at random, e.g.

  17. f r e q ( A B ) = f r e q ( A ) f r e q ( B ) f r e q ( A B C ) = f r e q ( A ) f r e q ( B ) f r e q ( C ) f r e q ( A B ) 6 = f r e q ( A ) f r e q ( B ) D = f r e q ( A B ) ° f r e q ( A ) f r e q ( B ) A B Linkage disequilibrium At LE, alleles in gametes are independentof each other: When linkage disequilibrium (LD) present, alleles are no longer independent --- knowing that one allele is in the gamete provides information on alleles at other loci The disequilibrium between alleles A and B is given by

  18. Forces that Generate LD • Selection • Drift • Migration (admixture) • Mutation • Population structure (stratification)

  19. f r e q ( A B ) = f r e q ( A ) f r e q ( B ) + D A B Departure from LE LE value t D ( t ) = D ( 0 ) ( 1 ° c ) Initial LD value The Decay of Linkage Disequilibrium The frequency of the AB gamete is given by If recombination frequency between the A and B loci is c, the disequilibrium in generation t is Note that D(t) -> zero, although the approach can be slow when c is very small Not surprising that very tightly-linked markers (c <<0.01) are often in LD

  20. 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

  21. Quantitative Genetics The analysis of traits whose variation is determined by both a number of genes and environmental factors Phenotype is highly uninformative as to underlying genotype

  22. 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

  23. Consider a specific locus influencing the trait Phenotypic distribution of a trait For this locus, mean phenotype = 0.15, while overall mean phenotype =0

  24. 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

  25. 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 alleles D and d, where allele D significantly increases your disease risk In particular, Pr(disease | DD) = 0.5, so that the Penetrance of genotype DD is 50% Suppose Pr(disease | Dd ) = 0.2, Pr(disease | dd) = 0.05 dd individuals can rarely display the disease, largely because of exposure to adverse environmental conditions

  26. dd individuals can give rise to phenocopies 5% of the time, showing the disease but not as a result of carrying the risk allele 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 %) Pr(Dd | disease) = 0.442, Pr(dd | disease) = 0.497

  27. Genotypic value Phenotypic value -- we will occasionally also use z for this value Environmental value Basic model of Quantitative Genetics Basic model: P = G + E G = average phenotypic value for that genotype if we are able to replicate it over the universe of environmental values, G = E[P] G x E interaction --- G values are different across environments. Basic model now becomes P = G + E + GE

  28. C C C + a(1+k) C + a + d C + 2a C + 2a C -a C + d C + a 2a = G(Q2Q2) - G(Q1Q1) Contribution of a locus to a trait Q1Q1 Q2Q1 Q2Q2 d measures dominance, with d = 0 if the heterozygote is exactly intermediate to the two homozygotes d = ak =G(Q1Q2 ) - [G(Q2Q2) + G(Q1Q1) ]/2 k = d/a is a scaled measure of the dominance

  29. Example: Apolipoprotein E & Alzheimer’s 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

  30. G = π + Æ + Æ + ± i j G i j i j The genotypic value predicted from the individual allelic effects is thus X Dominance deviations --- the difference (for genotype AiAj) between the genotypic value predicted from the two single alleles and the actual genotypic value, Average contribution to genotypic value for allele i Mean value, with b π = G ¢ f r e q ( Q Q ) G = π + Æ + Æ G i j i j i j G i j b G ° G = ± i j i j i j Fisher’s (1918) 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. Consider the genotypic value Gij resulting from an AiAj individual Since parents pass along single alleles to their offspring, the ai (the average effect of allele i) represent these contributions

  31. G = π + Æ + Æ + ± i j G i j i j Residual error Predicted value G = π + 2 Æ + ( Æ ° Æ ) N + ± i j G i j 1 2 1 Intercept Independent (predictor) variable N = # of Q2 alleles Regression slope Regression residual 8 2 Æ f o r N = 0 ; e . g , Q Q > 1 1 1 < 2 Æ + ( Æ ° Æ ) N = Æ + Æ f o r N = 1 ; e . g , Q Q 1 2 1 1 1 1 2 > : 2 Æ f o r N = 2 ; e . g , Q Q 1 2 2 Fisher’s decomposition is a Regression A notational change clearly shows this is a regression,

  32. Allele Q1 common, a2 > a1 Allele Q2 common, a1 > a2 Both Q1 and Q2 frequent, a1 = a2 = 0 Slope = a2 - a1 G21 G22 G G11 0 1 2 N

  33. Mean π = 2 p a ( 1 + p k ) G 2 1 Allelic effects Æ = p a [ 1 + k ( p ° p ) ] 2 1 1 2 Æ = ° p a [ 1 + k ( p ° p ) ] 1 2 1 2 Dominance deviations ± = G ° π ° Æ ° Æ i j i j G i j Consider a diallelic locus, where p1 = freq(Q1)

  34. n ≥ ¥ ) ( X k k ( ) ( ) A = Æ + Æ i k k = 1 A ( G ) = Æ + Æ i j i j Average effects and Additive Genetic Values The a values are the average effects of an allele A key concept is the Additive Genetic Value (A) of an individual Why all the fuss over A? Suppose father has A = 10 and mother has A = -2 for (say) blood pressure KEY: parents only pass single alleles to their offspring. Hence, they only pass along the A part of their genotypic Value G Expected blood pressure in their offspring is (10-2)/2 = 4 units above the population mean. Offspring A = Average of parental A’s

  35. G = π + ( Æ + Æ ) + ± i j g i j i j 2 2 2 2 n n æ ( G ) = æ ( π + ( Æ + Æ ) + ± ) = æ ( Æ + Æ ) + æ ( ± ) X X g i j i j i j i j k k k ( ) ( ) ( ) 2 2 2 æ ( G ) = æ ( Æ + Æ ) + æ ( ± ) As Cov(a,d) = 0 i j i j k k = 1 = 1 2 2 2 æ = æ + æ G A D Dominance Genetic Variance (or simply dominance variance) Additive Genetic Variance (or simply Additive Variance) Genetic Variances

  36. Key concepts (so far) • ai = 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)

  37. Q1Q1 Q1Q2 Q2Q2 m X 2 2 2 æ = 2 E [ Æ ] = 2 Æ p 0 a(1+k) 2a i A i i = 1 2 2 2 æ = 2 p p a [ 1 + k ( p ° p ) ] 1 2 1 2 A When dominance present, asymmetric function of allele frequencies Dominance effects additive variance m m X X 2 2 2 æ = 2 E [ ± ] = ± p p i j D i j i j = 1 = 1 2 2 æ = ( 2 p p a k ) 1 2 D Equals zero if k = 0 This is a symmetric function of allele frequencies Since E[a] = 0, Var(a) = E[(a -ma)2] = E[a2] One locus, 2 alleles: One locus, 2 alleles:

  38. VA Allele frequency, p Additive variance, VA, with no dominance (k = 0)

  39. Complete dominance (k = 1) VA VD Allele frequency, p

  40. G = π + ( Æ + Æ + Æ + Æ ) + ( ± + ± ) i j k l G i j k l i j k j + ( Æ Æ + Æ Æ + Æ Æ + Æ Æ ) i k i l j k j l + ( Æ ± + Æ ± + Æ ± + Æ ± ) i k l j k l k i j l i j + ( ± ± ) i j k l = π + A + D + A A + A D + DD G Additive x Additive interactions -- interactions between a single allele at one locus with a single allele at another Dominance x dominance interaction --- the interaction between the dominance deviation at one locus with the dominance deviation at another. Additive x Dominant interactions -- interactions between an allele at one locus with the genotype at another, e.g. allele Ai and genotype Bkj Dominance value -- interaction between the two alleles at a locus Additive Genetic value 2 2 2 2 2 2 æ = æ + æ + æ + æ + æ G A D A A A D D D Epistasis These components are defined to be uncorrelated, (or orthogonal), so that

  41. 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

  42. 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

  43. One allele IBD Both alleles IBD No alleles IBD Genetic Covariance between relatives 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. Genetic covariances arise because two related individuals are more likely to share alleles than are two unrelated individuals.

  44. G = A + D = Æ + Æ + D p p p x x 1 1 G = A + D = Æ + Æ + D o o o y y 1 1 IBD allele Non-IBD alleles Parent-offspring genetic covariance Cov(Gp, Go) --- Parents and offspring share EXACTLY one allele IBD Denote this common allele by A1

  45. o v ( ; G ) = C o v ( Æ + Æ + D ; Æ + Æ + D C G o p x x y y 1 1 1 1 = C o v ( Æ ; Æ ) + C o v ( Æ ; Æ ) + C o v ( Æ ; D ) y y 1 1 1 1 1 + C o v ( Æ ; Æ ) + C o v ( Æ ; Æ ) + C o v ( Æ ; D ) x x y x y 1 1 + C o v ( D ; Æ ) + C o v ( D ; Æ ) + C o v ( D ; D ) x x y x y 1 1 1 1 1 • By construction, a and D are uncorrelated • By construction, a from non-IBD alleles are uncorrelated • By construction, D values are uncorrelated unless both alleles are IBD All white covariance terms are zero.

  46. Ω 0 i f x = 6 y ; i . e . , n o t I B D C o v ( Æ ; Æ ) = x y V a r ( A ) = 2 i f x = y ; i . e . , I B D V a r ( A ) = V a r ( Æ + Æ ) = 2 V a r ( Æ ) 1 2 1 s o t h a t V a r ( Æ ) = C o v ( Æ ; Æ ) = V a r ( A ) = 2 1 1 1 Hence, relatives sharing one allele IBD have a genetic covariance of Var(A)/2 The resulting parent-offspring genetic covariance becomes Cov(Gp,Go) = Var(A)/2

  47. The half-sibs share one allele IBD • occurs with probability 1/2 The half-sibs share no alleles IBD • occurs with probability 1/2 Half-sibs 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

  48. Paternal 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 Full-sibs Each sib gets exact one allele from each parent Prob(exactly one allele IBD) = 1/2 = 1- Prob(0 IBD) - Prob(2 IBD)

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